System of Particles and Rotational Motion Test - 14

Centre of mass of A bangle is at centre.

If disc rotates with constant angular velocity, then angular acceleration of the disc is zero, i.e.,

\(\alpha={\Delta \omega\over \Delta t}=0\)

Density, \(\rho (x)=a(1+bx^2)\), at b = 0,

\(\rho\)(x) = a = constant

centre of mass, at x = 0.5 m

(a) \({3\over4}\times {2\over3}={1\over2}=0.5m\)

(b) \({4\over3}\times {2\over3}\ne0.5m\)

(c) \({3\over4}\times {3\over2}\ne 0.5m\)

(d) \({4\over3}\times {3\over2}\ne0.5m\)

(a) For a general rotational motion where axis of rotation is not symmetric, angular momentum z and angular velocity \(\omega\) need not to be parallel.

(c) \(\vec p=m\vec v\) so \(\vec p|| \vec v\)

(a) \(\tau=Fr\,sin\theta=0\)   \(\because\, \vec F|| \vec r\)

(b) \(\tau=Fr\,sin\theta=0\)   \(\because\, \vec \tau|| \vec F\)

(c) \(\tau=Fr\,sin\theta=0\)   \(\because\, \vec \tau|| \vec F\)

(d) \(\tau=Fr\,sin\theta=0\) x \(r\, sin \theta=0\)

(a) False

Frictional force acts opposite to the direction of motion of the center of mass of a body. In the case of rolling, the direction of motion of the center of mass is backward. Hence, frictional force acts in the forward direction.

(b) True

Rolling can be considered as the rotation of a body about an axis passing through the point of contact of the body with the ground. Hence, its instantaneous speed is zero.

(c) False

When a body is rolling, its instantaneous acceleration is not equal to zero. It has some value.

(d) True

When perfect rolling begins, the frictional force acting at the lowermost point becomes zero. Hence, the work done against friction is also zero.

(e) True

The rolling of a body occurs when a frictional force acts between the body and the surface. This frictional force provides the torque necessary for rolling. In the absence of a frictional force, the body slips from the inclined plane under the effect of its own weight.

As no torque is exerted by the person jumping, radially away from the centre of the round, let the total moment of inertia of the system is 2I and the round is revolving with angular speed w. Since the angular momentum of the person when it jumps off the round is Iw the actual momentum of round seen from ground is 2I\(\omega\) – I\(\omega\) = I\(\omega\)

So, we conclude that the angular speed remains same i.e., \(\omega\).