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System of Particles and Rotational Motion

Multiple Choice Questions

61.

A wheel of radius 0.4 m can rotate freely about its axis as shown in the figure. A string is wrapped over its rim and a mass of 4 kg is hung. An angular acceleration of 8 rad-s^-2 produced in it due to the torque. Then, moment of inertia of the wheel is (g =10 ms^-2 )

2 kg-m²
1 kg-m²
4 kg-m²
8 kg-m²

62.

An object start sliding on a frictionless inclined plane and from same height another object start falling freely.

both will reach with same speed
both will reach with same acceleration
both will reach in same time
None of the above

63.

Two rigid bodies A and B rotate with rotational kinetic energies E, and E, respectively. The moments of inertia of A and B about the axis of rotation are I_A and I_B respectively.

If $I_{A} = \frac{I_{B}}{2}$ and E_A = 100 = E_B, the ratio of angular momentum (L_A ) of A to the angular momentum ( L_B ) of B is

64.

The working principle of a ball point pen is

Bernoulli's theorem
surface tension
gravity
viscosity

65.

Progressive waves are represented by the equation y₁ = a sin (ωt - x ) and y₂ = bcos (ωt - x ). The phase difference wave is

0^o
45^o
90^o
180^o

66.

If applied torque on a system is zero, i.e.,Τ = 0 , then for that system

ω = 0
α = 0
J = 0
F = 0

67.

Match the following

Angular momentum	1. [M^-1 L² T^-2 ]
B. Torque	2 [M¹ L² T^-2
C. Gravitational constant	3.[M¹ L² T^-2]
D. Tension	4.[M¹ L² T^-1]

C- 2, D - 1
A - 4, B - 3
A - 3, C -2
B-2, A - 1

68.

A tangential force acting on the top of sphere of mass m kept on a rough horizontal place as shown in figure

If the sphere rolls without slipping, then the acceleration with which the centre of sphere moves, is

$\frac{10 F}{7 m}$
$\frac{F}{2 m}$
$\frac{3 F}{7 m}$
$\frac{7 F}{2 m}$

69.

A solid sphere is set into motion on a rough horizontal surface with a linear speed v in the forward direction and an angular speed $\frac{v}{R}$ in the anticlockwise direction as shown in figure. Find the linear speed of the sphere when it stops rotating and ω = $\frac{v}{R}$

$\frac{3 v}{5}$
$\frac{2 v}{5}$
$\frac{4 v}{3}$
$\frac{7 v}{3}$

70.
Two blocks of masses m₁ and m₂ are connected by a spring of spring constant k. The block of mass m₂ is given a sharp empulse so that it acquires a velocity v₀towards right. Find the maximum elongation that the spring will suffer.

${[\frac{m_{1} m_{2}}{m_{1} + m_{2}}]}^{\frac{1}{2}}$ v₀

$(\frac{m_{1} + m_{2}}{m_{1} - m_{2}})$ v₀

${[\frac{m_{1} + m_{2}}{m_{1} - m_{2}}]}^{\frac{1}{2}}$ v₀

${[\frac{2 m_{1} + m_{2}}{m_{1} m_{2}}]}^{\frac{1}{2}}$ v₀

${[\frac{m_{1} m_{2}}{m_{1} + m_{2}}]}^{\frac{1}{2}}$ v₀

The centre of mass is the location of particles within a system where the total mass of the system can be considered concentrated. When the system of particles is moving, the center of mass moves along with it.

The centre of mass of velocity equation is the sum of each particle's momentum ( mass times velocity ) divided by the total mass of the system.

The velocity of the centre of mass of two particles

v_cm = $\frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}$

When v₁ =0 and v₂ =v₀, then

$v_{cm} = \frac{m_{2} v_{0}}{m_{1} + m_{2}}$

Now, let 'x' be the elongation in the spring.

Change in potential energy = potential energy stored in spring

⇒ $\frac{1}{2} m_{2} v_{0}^{2} - \frac{1}{2} (m_{1} + m_{2})$ ${(\frac{m_{2} v_{0}}{m_{1} + m_{2}})}^{2} = \frac{1}{2} {kx}^{2}$

⇒ m₂ $v_{0}^{2}$ $\{1 - \frac{m_{2}}{m_{1} + m_{2}}\}$ = kx²

⇒ $m_{2} v_{0}^{2} [\frac{m_{1} + m_{2} - m_{2}}{m_{1} + m_{2}}]$ = kx²

This gives

x = ${(\frac{m_{1} m_{2}}{m_{1} + m_{2}})}^{\frac{1}{2}}$ v₀