## A mass m supported by a massless string wound around a uniform hollow cylinder of mass m and radius R…

Q: A mass m supported by a massless string wound around a uniform hollow cylinder of mass m and radius R. If the string does not slip on the cylinder, with what acceleration will the mass fall on release?

(a)2g/3

(b)g/2

(c)5g/6

(d)g

Ans: (b)

Sol: Applying Laws of Motion

$\large mg-T = ma$

$\large mg-T = m (R \alpha)$ …(i)

Taking Torque about centre of pulley

$\large T\times R = mR^2 \alpha$ …(ii)

Solving (i) & (ii)

a = g/2

## A small mass m is attached to a massless string whose other end is fixed at P as shown in the figure…

Q: A small mass m is attached to a massless string whose other end is fixed at P as shown in the figure. The mass is undergoing circular motion in the x-y plane with centre at O and constant angular speed ω. If the angular momentum of the system, calculated about O and P are denoted by LO and LP respectively, then

(a)Lo and LP do not vary with time

(b)Lo varies with time while LP remains constant

(c)Lo remains constant while LP varies with time

(d)Lo and LP both vary with time

Ans: (c)

SOl: $\large \vec{L} = m(\vec{r}\times \vec{v})$

## A small object uniform density rolls up a curved surface with an initial velocity v. It reaches up to a maximum height…

Q: A small object uniform density rolls up a curved surface with an initial velocity v. It reaches up to a maximum height of 3v2/4g with respect to the initial position. The object is

(a)ring

(b)solid sphere

(c)hollow sphere

(d)disc

Ans: (d)

Sol: Applying energy conservation

$\large \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = m g h$ ; (where I = Moment of Inertia of the object)

$\large \frac{1}{2}mv^2 + \frac{1}{2}I(v/R)^2 = m g (\frac{3v^2}{4g})$

$\large I = \frac{1}{2}mR^2$

Hence object is disc

## A ball moves over a fixed track as shown in the figure. From A to B the ball rolls without slipping if surface…

Q: A ball moves over a fixed track as shown in the figure. From A to B the ball rolls without slipping if surface BC is frictionless and KA, KB and KC are kinetic energies of the ball at A, B and C respectively, then

(a)hA>hc; KB>KC

(b)hA>hc; KC>KA

(c)hA = hc; KB = KC

(d)hA<hc; KB>KC

Ans: (a)

Sol: On frictionless part BC , there is Zero torque , therefore angular velocity and hence rotational K.E remain constant . Whereas moving from B to C Translational K.E is converted into gravitational P.E .

## A solid sphere of radius R has moment of inertia I about its geometrical axis. It is melted into a disc…

Q: A solid sphere of radius R has moment of inertia I about its geometrical axis. It is melted into a disc of radius r and thickness t. If it’s moment of inertia about the tangential axis (which is perpendicular to plane of the disc), is also equal to I, then the value of r is equal to

(a) $\large \frac{2}{\sqrt{15}}R$

(b) $\large \frac{2}{\sqrt{5}}R$

(c) $\large \frac{3}{\sqrt{15}}R$

(d) $\large \frac{\sqrt{3}}{\sqrt{15}}R$

Ans: (a)

Sol: According to question;

$\large \frac{2}{5}MR^2 = \frac{1}{2}Mr^2 + M r^2$ ; (By applying parallel axes theorem)

$\large \frac{2}{5}MR^2 = \frac{3}{2}Mr^2$

$\large r = \frac{2}{\sqrt{15}}R$