Rotational Motion

A uniform disk of mass 300 kg is rotating freely about a vertical axis through its centre with constant angular velocity ωo ….

by

Q: A uniform disk of mass 300 kg is rotating freely about a vertical axis through its centre with constant angular velocity ωo . A boy of mass 30 kg starts from the centre and moves along a radius to the edge of the disk. The angular velocity of the disk now is

(A) $\frac{\omega_0}{6}$

(B) $\frac{\omega_0}{5}$

(C) $\frac{4\omega_0}{5}$

(D) $\frac{5 \omega_0}{6}$

Click to See Solution :
Ans: (D)
Sol: As Torque = 0 ; angular momentum remains conserved

$\displaystyle L = (0 + \frac{300R^2}{2})\omega_0 = (\frac{300R^2}{2} + 30 R^2)\omega $

$\displaystyle 150 \omega_0 = 180 \omega $

$\displaystyle \omega = \frac{5}{6}\omega_0 $

 

A sphere rolls without sliding on a rough inclined plane (only mg and contact forces are acting on the body)….

by

Q: A sphere rolls without sliding on a rough inclined plane (only mg and contact forces are acting on the body). The angular momentum of the body:

(A) about centre is conserved

(B) is conserved about the point of contact

(C) is conserved about a point whose distance from the inclined plane is greater than the radius of the sphere

(D) is not conserved about any point

Click to See Solution :
Ans: (C)
Sol: Angular momentum will be conserved if the net
torque is zero . Now for the sphere to move down:
mg sinθ > μ mg cosθ

Numerical

Let x be the perpendicular distance of the point
(as shown in figure) about which torque remains
zero.
for τ = 0 ; x > R as shown

Numerical

Note: As mgsinθ > μ mgcosθ , the point should be inside the sphere .

 

A ring of mass m and radius R rolls on a horizontal rough surface without slipping due to an applied force F ….

by

Q: A ring of mass m and radius R rolls on a horizontal rough surface without slipping due to an applied force F . The friction force acting on ring is :

Numerical

(A) F/3

(B) 2F/3

(C) F/4

(D) Zero

Click to See Solution :
Ans: (D)
Sol: F + f = m a …. (1)

Also ; F R – f R = I a/ R

F – f = m a …. (2)

[I = m R2]

From (1) & (2)

f = 0

 

Moment of inertia of a uniform quarter disc of radius R and mass M about an axis through its centre of mass ….

by

Q: Moment of inertia of a uniform quarter disc of radius R and mass M about an axis through its centre of mass and perpendicular to its plane is :

(A) $\frac{MR^2}{2} – M(\frac{4R}{3\pi})^2$

(B) $\frac{MR^2}{2} – M(\sqrt{2} \frac{4R}{3\pi})^2$

(C) $\frac{MR^2}{2} + M(\frac{4R}{3\pi})^2$

(D) $\frac{MR^2}{2} + M(\sqrt{2} \frac{4R}{3\pi})^2$

Click to See Solution :
Ans: (B)

Sol: Numerical

Moment of Inertia about ‘O’ is $I = \frac{MR^2}{2}$

By parallel-axis theorem

$\frac{MR^2}{2} = I_{cm} + M(\sqrt{2} \frac{4R}{3\pi})^2 $

$I_{cm} = \frac{MR^2}{2} – M(\sqrt{2} \frac{4R}{3\pi})^2 $