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

Solutions of H.C. Verma’s Concepts of Physics chapter 7 (Circular Motion) are given below. You can download H.C. Verma Solutions in PDF format by simply clicking on pop-out button at right corner and download PDF or clicking on print command and save as PDF. 

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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).

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