HC Verma Concepts Of Physics Exercise Solutions Of Chapter 7 (Circular Motion)

HC Verma Concepts of Physics Solutions - Part 1, Chapter 7 - Circular Motion:

EXERCISE

1. Find the acceleration of the moon with respect to the earth from the following data: Distance between the earth and the moon = 3.85 * 10⁵ km and the time taken by the moon to complete one revolution around the earth = 27.3 days.

Sol:

Given: Distance between the earth and the moon, r = 3.85 * 10⁵ km = 3.85 * 10⁸ m; T = 27.3 days = 2.36 * 10⁶ s.

Angular speed, ω = 2π/T = 2.66 * 10^-6 rad/s.

Acceleration, a = ω²r = (2.66 * 10^-6)² * 3.85 * 10⁸ = 2.73 * 10^-3 m/s².

2. Find the acceleration of a particle placed on the surface of the earth at the equator due to earth's rotation. The diameter of earth = 12800 km and it takes 24 hours for the earth to complete one revolution about its axis.

Sol:

Given: Distance between the centre of earth and the particle, r = 12800/2 km = 64 * 10⁵ m; T = 24 hrs. = 86400 s.

Angular speed, ω = 2π/T = 7.27 * 10^-5 rad/s.

Acceleration, a = ω²r = (7.27 * 10^-5)² * 64 * 10⁵ = 0.0336 m/s².

3. A particle moves in a circle of radius 1.0 cm at a speed given by v = 2.0 t where v is in cm/s and t in seconds. (a) Find the radial acceleration of the particle at t = 1 s. (b) Find the tangential acceleration at t = 1 s. (c) Find the magnitude of the acceleration at t = 1 s.

Sol:  

Given: radius, r = 1.0 cm; v = 2.0t.

Speed of particle at 1 s, v_{t = 1 s} = 2 cm/s

(a) The radial acceleration of the particle at t = 1 s

→ (a_r) _{t = 1 s} = (v_{t = 1 s})²/r = 4 cm/s².

(b) The tangential acceleration of the particle at t = 1 s 
→ (a_t) _{t = 1 s} = (dv/dt) _{t = 1 s} = d(2t)/dt = 2 cm/s².

(c) The magnitude of the acceleration at t = 1 s

→ a = √ (a_r² + a_t²) = √ (4² + 2²) = √20 cm/s².

4. A scooter weighing 150 kg together with its rider moving at 36 km/hr is to take a turn of radius 30 m. What horizontal force on the scooter is needed to make the turn possible?

Sol:

Given: speed, v = 36 km/hr = 10 m/s; total mass, M = 150 kg; radius of a turn, r = 30 m.

To make the turn possible, horizontal force (F) must be equal to Centripetal force (F_c). Otherwise there will be skidding.

We know, Centripetal force (F_c) = mv²/r = (150 * 10²)/30 = 500 N.

Therefore, horizontal force (F) = 500 N.

5. If the horizontal force needed for the turn in the previous problem is to be supplied by the normal force by the road, what should be the proper angle of banking?

Sol: 

Given: speed, v = 36 km/hr = 10 m/s; total mass, M = 150 kg; radius of a turn, r = 30 m.

From the diagram

R cos θ = Mg ------------------ (1)

R sin θ = F_C

Or, R sin θ = Mv²/r ---------- (2)

Dividing equation (2) by equation (1)

We get, tan θ = v²/rg = 10²/ (30 * 10) = 1/3

Or, θ = tan^-1 (1/3).

6. A park has a radius of 10 m. If a vehicle goes round it at an average speed of 18 km/hr, what should be the proper angle of banking?

Sol: 

Given: speed, v = 18 km/hr = 5 m/s; radius of a turn, r = 10 m.

Angle of banking is given by,

→ tan θ = v²/rg = 5²/ (10 * 10) = ¼

Or, θ = tan^-1 (1/4).

7. If the road of the previous problem is horizontal (no banking), what should be the minimum friction coefficient so that a scooter going at 18 km/hr does not skid?

Sol: 

Given: speed, v = 18 km/hr = 5 m/s; radius of a turn, r = 10 m. 

Vertical motion:

→ R = Mg and

Horizontal motion:

→ Mv²/r = μR

Or, Mv²/r = μMg

Or, μ = v²/rg = 5²/ (10*10) = 0.25.

8. A circular road of radius 50 m has the angle of banking equal to 30°. At what speed should a vehicle go on this road so that the friction is not used?

Sol: 

Given: θ = 30⁰; radius of a turn, r = 50 m.

Angle of banking is given by,

→ tan θ = v²/rg

Or, tan 30⁰ = v²/ (50 * 10)

Or, v = √ (500 * tan 30⁰) = 17 m/s.

9. In the Bohr model of hydrogen atom, the electron is treated as a particle going in a circle with the centre at the proton. The proton itself is assumed to be fixed in an inertial frame. The centripetal force is provided by the Coloumb attraction. In the ground state, the electron goes round the proton in a circle of radius 5.3*10^-11 m. Find the speed of the electron in the ground state. Mass of the electron = 9.1 * 10^-31 kg and charge of the electron = 1.6 * 10^-19 C.

Sol: 

Given: q₁ = q₂ = q = 1.6 * 10^-19 C; radius of circle, r = 5.3*10^-11 m; Mass of the electron, M = 9.1 * 10^-31 kg.

The centripetal force is provided by the Coloumb attraction.

Therefore, Mv²/r = kq₁q₂/r²

Or, v = √ (kq²/Mr)         [k = 9 * 10⁹]

Or, v = √ [(9 * 10⁹) * (1.6 * 10^-19)²/ (9.1 * 10^-31) (5.3*10^-11)]

Or, v = √ (4.78 * 10¹²) = 2.2 * 10⁶ m/s.

10. A stone is fastened to one end of a string and is whirled in a vertical circle of radius R. Find the minimum speed the stone can have at the highest point of the circle.

Sol:

The centripetal force is provided by the gravitational attraction and tension force when the stone at highest point.

Therefore, Mv²/R = T + Mg

For the minimum speed, T = 0

So, Mv²/R = Mg

Or, v = √ (Rg).

Therefore, the minimum speed the stone is √ (Rg).

11. A ceiling fan has a diameter (of the circle through the outer edges of the three blades) of 120 cm and rpm 1500 at full speed. Consider a particle of mass 1 g sticking at the outer end of a blade. How much force does it experience when the fan runs at full speed? Who exerts this force on the particle? How much force does the particle exert on the blade along its surface?

Sol:

Given: diameter of fun = 120 cm; radius, r = 60 cm = 0.6 m; N = 1500 rpm; angular speed, ω = 2πN/60 = 50π rad/s; mass of particle, M = 1 g = 0.001 kg.

Force on the particle, F = M ω²r

Or, F = 0.001 * (50π)² * 0.6 = 14.8 N.

From Newton’s 3^rd law, particle exert equal and opposite force on the blade along its surface.

So, force on the blade = 14.8 N.

12. A mosquito is sitting on an L.P. record disc rotating on a turn table at 33⅓ revolutions per minute. The distance of the mosquito from the centre of the turn table is 10 cm. Show that the friction coefficient between the record and the mosquito is greater than π²/81. Take; g = 10 m/s².

Sol:

Given: radius, r = 10 cm = 0.1 m; N = 33⅓ rpm; angular speed, ω = 2πN/60 = 10π/9 rad/s; Let mass of mosquito be M.

For equilibrium of mosquito

Frictional force ≥ centripetal force

Or, μMg ≥ Mω²r

Or, μ ≥ ω²r/g = π²/81

Therefore, the friction coefficient between the record and the mosquito is greater than π²/81.

13. A simple pendulum is suspended from the ceiling of a car taking a turn of radius 10 m at a speed of 36 km/h. Find the angle made by the string of the pendulum with the vertical if this angle does not change during the turn. Take g = 10 m/s². 

Sol:

Given: radius, r = 10 m; speed, v = 36 km/h = 10 m/s; g = 10 m/s².

A pendulum is suspended from the ceiling of a car taking a turn.

From the diagram:

T cos θ = mg ------------- (1)

T sin θ = mv²/r ----------- (2)

Dividing equation (2) by equation (1)

We get, tan θ = v²/rg = 1

Or, θ = 45⁰.

14. The bob of a simple pendulum of length 1 m has mass 100 g and a speed of 1.4 m/s at the lowest point in its path. Find the tension in the string at this instant.

Sol:

Given: Length of pendulum, L = 1 m; mass of bob, M = 100 g = 0.1 kg; speed, v = 1.4 m/s.

Centripetal acceleration, a_c = v²/L

Or, a_c = 1.4²/1 = 1.96 m/s².

             M a_c = T – Mg

Or, 0.1 * 1.96 = T – 0.1 * 10

Or, T = 1.2 N

So, the tension in the string at this instant is 1.2 N.

15. Suppose the bob of the previous problem has a speed of 1.4 m/s when the string makes an angle of 0.20 radian with the vertical. Find the tension at this instant. You can use cos θ = 1 – θ²/2 and sin θ ≈ 0 for small θ.

Sol:

Given: Length of pendulum, L = 1 m; mass of bob, M = 100 g = 0.1 kg; speed, v = 1.4 m/s; θ = 0.20 radian; cos θ = 1 – θ²/2 and sin θ ≈ 0;

Centripetal acceleration, a_c = v²/L

Or, a_c = 1.4²/1 = 1.96 m/s².

Mg sin θ ≈ 0 and

         Ma_c = T – Mg cos θ

Or, T = M a_c + Mg (1 – θ²/2)

Or, T = 0.1 * 1.96 + 0.1 * 10 * (1 – 0.2²/2)

Or, T = 1.16 N

So, the tension in the string at this instant is 1.16 N.

16. Suppose the amplitude of a simple pendulum having a bob of mass m is θ₀. Find the tension in the string when the bob is at its extreme position.

Sol:

At the extreme position, velocity of the pendulum is zero.

So, there is no centrifugal force.

So, T = mg cos θ₀.

17. A person stands on a spring balance at the equator, (a) by what fraction is the balance reading less than his true weight? (b) If the speed of earth's rotation is increased by such an amount that the balance reading is half the true weight, what will be the length of the day in this case?

Sol:

Given: radius of earth, r = 6400 * 10³ m; ω = 2π/T = 2π/ (24 *3600); g = 10 m/s².

Centripetal acceleration, a_c = ω²r

(a) True weight = Mg

From the diagram

We get, Ma_c = Mg – R

New weight = R = Mg – M ω²r

Fraction of reading less than his true weight is

= (True weight – New weight)/ (True weight)

= [Mg – (Mg – M ω2r)]/ (Mg) = ω2r/g

= [2π/ (24 *3600)]² * 6400 * 10³ /10

= 3.5 * 10^-3.

(b) New weight = ½ * true weight

From the diagram

We get, Ma_c = Mg – R

Or, M ω²r = Mg – R

Or, ω²r = g/2       [R = ½ Mg]

Or, ω = √ (g/2r)

Or, T = 2π/√ (g/2r)

Or, T = 2π/√ [10/ (2*6400*10³)] = 7108 s ≈ 2 hours.

18. A turn of radius 20 m is banked for the vehicles going at a speed of 36 km/h. If the coefficient of static friction between the road and the tyre is 0.4, what are the possible speeds of a vehicle so that it neither slips down nor skids up?

Sol:

Given, v = 36 km/hr = 10 m/s; r = 20 m, μ = 0.4.

The road is banked with an angle,

→ θ = tan^-1 (v²/rg) 

Or, θ = tan^-1 (1/2) = 26.6⁰.

Or, tan θ = 0.5

When the car travels at max. Speed (v₁) so that it skids upward, f = μR acts downward as shown in Fig.1 and a_c = v₁²/r.

R cos θ – Mg – μR sin θ = 0

Or, R = Mg/ (cos θ – μ sin θ) ---------- (1)

Ma_c = R sin θ – μR cos θ

Or, Mv₁²/r = R (sin θ + μ cos θ) ------- (2)

Solving equation (1) and (2)

We get, v₁ = √ [rg (sin θ + μ cos θ) / (cos θ – μ sin θ)]

Or, v₁ = √ [rg (tan θ + μ) / (1 – μ tan θ)]

Or, v₁ = √ [20*10 (0.5 + 0.4) / (1 – 0.4* 0.5)]

Or, v₁ = 15 m/s = 54 km/hr

When the car travels at min. Speed (v₂) so that it slips downward, f = μR acts upward as shown in Fig.2 and a_c = v₂²/r.

R cos θ – Mg + μR sin θ = 0

Or, R = Mg/ (cos θ + μ sin θ) ---------- (1)

Ma_c = R sin θ – μR cos θ

Or, Mv₂²/r = R (sin θ – μ cos θ) ------- (2)

Solving equation (1) and (2)

We get, v₂ = √ [rg (sin θ – μ cos θ) / (cos θ + μ sin θ)]

Or, v₂ = √ [rg (tan θ – μ) / (1 + μ tan θ)]

Or, v₂ = √ [20*10 (0.5 – 0.4) / (1 + 0.4* 0.5)]

Or, v₂ = 4.08 m/s = 14.7 km/hr

So, the possible speeds are between 14.7 km/hr and 54 km/hr.

19. A motorcycle has to move with a constant speed on an overbridge which is in the form of a circular arc of radius R and has a total length L. Suppose the motorcycle starts from the highest point, (a) What can its maximum velocity be for which the contact with the road is not broken at the highest point ? (b) If the motorcycle goes at speed 1/√2 times the maximum found in part (a), where will it lose the contact with the road? (c) What maximum uniform speed can it maintain on the bridge if it does not lose contact anywhere on the bridge?

Sol:

R = radius of the bridge; L = total length of the over bridge; N = normal reaction. Centripetal acceleration, a_c = v²/R

(a) At the height point:

Writing Newton’s 2^nd law of motion along vertical direction:

→ Ma_c = Mg – N

Or, Mv²/R = Mg – N

For the maximum velocity, N = 0

Therefore, M (v_max)²/R = Mg

Or, v_max = √ (Rg).

(b) Given: v = 1/√2 [√ (Rg)]

Suppose it loses contact at B. N = 0

Writing Newton’s 2^nd law of motion along OB direction:

→ Ma_c = Mg cos θ – N

Or, Mv²/R = Mg cos θ

Or, θ = cos^-1 (Rg/2Rg) = cos^-1 (1/2) = π/3

So, the distance from the height point, where it will lose the contact is πL/3.

(c) Let the uniform speed on the bridge be v.

The chances of losing contact is maximum at the end of the bridge for which α = L/2R and N = 0.

Writing Newton’s 2^nd law of motion along OC direction:

→ Ma_c = Mg cos α – N

Or, Mv²/R = Mg cos α

Or, v = √ [Rg cos (L/2R)].

20. A car goes on a horizontal circular road of radius R, the speed increasing at a constant rate dv/dt = a. The friction coefficient between the road and the tyre is μ. Find the speed at which the car will skid.

Sol:

Given: radius = R; coefficient of friction = μ

Since the motion is nonuniform, the acceleration has both radial & tangential component.

So, a_t = a and a_r = v²/R

Resultant acceleration, a_R = √ [(a_r)² + (a_t)²]

Or, a_R = √ [(v²/R)² + (a)²]

Inertia force is provided by frictional force.

Therefore, M a_R = μMg

Or, √ [(v²/R)² + (a)²] = μg

Or, v = [(μ²g² – a²) R²]^1/4 

So, the speed at which the car will skid is [(μ²g² – a²) R²]^1/4.

21. A block of mass m is kept on a horizontal ruler. The friction coefficient between the ruler and the block is μ. The ruler is fixed at one end and the block is at a distance L from the fixed end. The ruler is rotated about the fixed end in the horizontal plane through the fixed end. (a) What can the maximum angular speed be for which the block does not slip? (b) If the angular speed of the ruler is uniformly increased from zero at an angular acceleration α, at what angular speed will the block slip?

Sol:

Given: friction coefficient = μ; radius of rotation = L.

(a) Uniform circular motion:

Let the angular speed be ω and a_c = ω²L.

Vertical motion of the block:

→ R = Mg

Horizontal motion of the block:

→ Ma_c = μR = μMg

Or, ω²L = μg

Or, ω = √ (μg/L)

So, the maximum angular speed is √ (μg/L).

(b) Uniformly accelerated circular motion:

Since the motion is nonuniform, the acceleration has both radial & tangential component.

Radial component of acceleration, a_r = ω²L and tangential component of acceleration, a_t = αL.

Resultant acceleration, a_R = √ [(a_r)² + (a_t)²]

Or, a_R = √ [(ω²L)² + (αL)²]

Inertia force is provided by frictional force.

Therefore, M a_R = μMg

Or, √ [(ω²L)² + (αL)²] = μg

Or, ω = [(μ²g² – α²L²)*1/L²]^1/4 

Or, ω = [(μg/L)² – α²]^1/4.

22. A track consists of two circular parts ABC and CDE of equal radius 100 m and joined smoothly as shown in figure (7-E1). Each part subtends a right angle at its centre. A cycle weighing 100 kg together with the rider travels at a constant speed of 18 km/h on the track, (a) Find the normal contact force by the road on the cycle when it is at B and at D. (b) Find the force of friction exerted by the track on the tyres when the cycle is at B, C and D. (c) Find the normal force between the road and the cycle just before and just after the cycle crosses C. (d) What should be the minimum friction coefficient between the road and the tyre, which will ensure that the cyclist can move with constant speed? Take g = 10 m/s².

Sol: 

Given: radius of curves, r = 100 m; total mass, M = 100 kg; speed, v = 18 km/hr = 5 m/s. 

(a) a_c = v²/r = 5²/100 = 0.25 m/s².

Writing Newton’s 2^nd law of motion at point B.

→ M a_c = Mg – R

Or, R = M (g - a_c) = 100 * (10 – 0.25)

Or, R = 975 N

Therefore, the normal contact force at B is 975 N.

Writing Newton’s 2^nd law of motion at point B.

→ M a_c = R – Mg 

Or, R = M (g + a_c) = 100 * (10 + 0.25)

Or, R = 1025 N

Therefore, the normal contact force at B is 1025 N.

(b) At B & D the cycle has no tendency to slide. So at B & D, frictional force is zero.

At ‘C’, angle made by the track with horizontal is θ = 45⁰.

Frictional force, f = Mg sin θ (uniform velocity)

Or, f = 100 * 10 * = 707 N.

(c) Before ‘C’ equation of motion of cycle and rider together.

→ M a_c = Mg cos θ – R

Or, R = M (g cos 45⁰ – a_c)

Or, R = 100 * (10*1/√2 – 0.25) = 682 N.

After ‘C’ equation of motion of cycle and rider together.

→ M a_c = R – Mg cos θ

Or, R = M (g cos 45⁰ + a_c)

Or, R = 100 * (10*1/√2 + 0.25) = 732 N.

(d) To find out the minimum desired coeff. of friction, we have to consider a point just before C. (where N is minimum)

Now, f = Mg sin θ

Or, μR = 100 * 10 * 1/√2

Or, μ = 707/682

So, μ = 1.037.

23. In a children's park a heavy rod is pivoted at the centre and is made to rotate about the pivot so that the rod always remains horizontal. Two kids hold the rod near the ends and thus rotate with the rod (figure 7-E2). Let the mass of each kid be 15 kg, the distance between the points of the rod where the two kids hold it be 3.0 m and suppose that the rod rotates at the rate of 20 revolutions per minute. Find the force of friction exerted by the rod on one of the kids.

Sol:

Given: mass of each kid, M = 15 kg; radius of rotation, r = 3/2 = 1.5 m; rotational speed, N = 20 rpm.

Angular speed, ω = 2πN/60 = 2π/3 rad/s

Frictional force, f = mω²r = 15 * (2π/3)² * 1.5 = 10π² 

So, the force of friction exerted by the rod on one of the kids is 10π².

24. A hemispherical bowl of radius R is rotated about its axis of symmetry which is kept vertical. A small block is kept in the bowl at a position where the radius makes an angle θ with the vertical. The block rotates with the bowl without any slipping. The friction coefficient between the block and the bowl surface is μ. Find the range of the angular speed for which the block will not slip.

Sol:

If the bowl rotates at maximum angular speed, the block tends to slip upwards. So, the frictional force acts downward.

Centripetal acceleration, a_c = (ω_max)² R sin θ and frictional force, f = μN

Writing Newton’s 1^st law along vertical direction:

→ N cos θ – Mg – f sin θ = 0

Or, N cos θ – Mg – μN sin θ = 0

Or, N = Mg/ (cos θ – μ sin θ)

Writing Newton’s 2^nd law along horizontal direction:

→ Ma_c = N sin θ + f cos θ

Or, M (ω_max)² R sin θ = N (sin θ + μ cos θ)

Or, M (ω_max)² R sin θ = Mg (sin θ + μ cos θ) / (cos θ – μ sin θ)
Or, ω_max = [g (sin θ + μ cos θ) / R sin θ (cos θ – μ sin θ)]^1/2.

If the bowl rotates at minimum angular speed, the block tends to slip downwards. So, the frictional force acts upward.

Centripetal acceleration, a_c = (ω_min)² R sin θ and frictional force, f = μN

Writing Newton’s 1^st law along vertical direction:

→ N cos θ – Mg + f sin θ = 0

Or, N cos θ – Mg + μN sin θ = 0

Or, N = Mg/ (cos θ + μ sin θ)

Writing Newton’s 2^nd law along horizontal direction:

→ Ma_c = N sin θ – f cos θ

Or, M (ω_min)² R sin θ = N (sin θ – μ cos θ)

Or, M (ω_min)² R sin θ = Mg (sin θ – μ cos θ) / (cos θ + μ sin θ)

Or, ω_min = [g (sin θ – μ cos θ) / R sin θ (cos θ + μ sin θ)]^1/2.

25. A particle is projected with a speed u at an angle θ with the horizontal. Consider a small part of its path near the highest position and take it approximately to be a circular arc. What is the radius of this circle? This radius is called the radius of curvature of the curve at the point.

Sol:

Let, the radius of curvature of the curve be r.

Velocity at the height point = u cos θ

Centripetal acceleration, a_c = u² cos² θ/r

Writing Newton’s 2^nd law along vertical direction

→ M a_c = Mg

Or, u² cos² θ/r = g

Or, r = u² cos² θ/g.

26. What is the radius of curvature of the parabola traced out by the projectile in the previous problem at a point where the particle velocity makes an angle θ/2 with the horizontal?

Sol:

Let ‘v’ the velocity at the pt where it makes an angle θ/2 with horizontal. The horizontal component remains unchanged.

Therefore, v cos θ/2 = u cos θ

Or, v = u cos θ / (cos θ/2)

Centripetal acceleration, a_c = [u cos θ / (cos θ/2)]²/r

Writing Newton’s 2^nd law along the radius

We get, Ma_c = Mg cos θ/2

Or, u² cos² θ / (cos² θ/2)r = g cos θ/2

Or, r = u² cos² θ / g (cos³ θ/2).

27. A block of mass m moves on a horizontal circle against the wall of a cylindrical room of radius R. The floor of the room on which the block moves is smooth but the friction coefficient between the wall and the block is μ. The block is given an initial speed v₀. As a function of the speed v write (a) the normal force by the wall on the block, (b) the frictional force by the wall and (c) the tangential acceleration of the block, (d) Integrate the tangential acceleration to obtain the speed [dv/dt = v(dv/ds)] of the block after one revolution.

Sol:

A block of mass ‘M’ moves on a horizontal circle against the wall of a cylindrical room of radius ‘R’. Friction coefficient between wall & the block is μ.

(a) The normal force by the wall on the block is Mv²/R.

(b) Frictional force by wall = μMv²/R.

(c) μMv²/R = Ma 
Or,  a = – μv²/R (Deceleration).

(d) dv/dt = v (dv/ds) = – μv²/R

Or, ds = – (R/μv) dv ------------ (1)

Integrating equation 1,

We get, s = – (R/μ) ln (v) + c

At s = 0, v = v₀ → c = (R/μ) ln (v₀)

So, s = (R/μ) ln (v₀/v) → (v/v₀) = e^{- μs/R}

For, one rotation s = 2πR

So, v = v₀ e^{- 2πμ}.

28. A table with smooth horizontal surface is fixed in a cabin that rotates with a uniform angular velocity ω in a circular path of radius R (figure 7-E3). A smooth groove AB of length L (<< R) is made on the surface of the table. The groove makes an angle θ with the radius OA of the circle in which the cabin rotates. A small particle is kept at the point A in the groove and is released to move along AB. Find the time taken by the particle to reach the point B.

Sol:

The cabin rotates with angular velocity ω & radius R

→ The particle experiences a force mRω².

The component of mRω² along the groove provides the required force to the particle to move along AB.

Therefore, mRω² cos θ = ma → a = Rω² cos θ

Length of groove = L

L = ut + ½ at² = 0*t + ½ (Rω² cos θ) t² 

Or, t = √ [2L/ (Rω² cos θ)].

29. A car moving at a speed of 36 km/hr is taking a turn on a circular road of radius 50 m. A small wooden plate is kept on the seat with its plane perpendicular to the radius of the circular road (figure 7-E4). A small block of mass 100 g is kept on the seat which rests against the plate. The friction coefficient between the block and the plate is μ = 0.58. (a) Find the normal contact force exerted by the plate on the block, (b) The plate is slowly turned so that the angle between the normal to the plate and the radius of the road slowly increases. Find the angle at which the block will just start sliding on the plate.

Sol:

Given: v = Velocity of car = 36 km/hr = 10 m/s; r = Radius of circular path = 50 m; M = mass of small body = 100 g = 0.1 kg; μ = Friction coefficient between plate & body = 0.58.

(a) The normal contact force exerted by the plate on the block

N = Mv²/r = (0.1 * 10²)/50 = 0.2 N.

(b) The plate is turned so the angle between the normal to the plate & the radius of the road slowly increases

N = Mv² cos θ/r ………………….. (1)

μN = Mv² sin θ/r ………………….. (2)

Dividing equation (2) by (1)

We get, tan θ = μ

Or, θ = tan^-1 (0.58) = 30⁰.

30. A table with smooth horizontal surface is placed in a cabin which moves in a circle of a large radius R (figure 7-E5). A smooth pulley of small radius is fastened to the table. Two masses m and 2m placed on the table are connected through a string going over the pulley Initially the masses are held by a person with the strings along the outward radius and then the system Is released from rest (with respect to the cabin). Find the magnitude of the initial acceleration of the masses as seen from the cabin and the tension in the string.

Sol:

Given: m₁ = m and m₂ = 2 m; let angular speed be ω.

Equation of motion of two blocks are following

1. m₁a = T – m₁ω²R

2. m₂a = m₂ω²R – T

Solving equation (1) and (2) and putting m₁ = m and m₂ = 2m

We get, a = ω²R/3 and T = (4mω²R)/3.

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