A uniformly thick wheel with moment of inertia I and radius R is free to rotate about its centre of mass (see fig). A massless string is wrapped over its rim and two blocks of masses $m_{1}$ and $m_{2}\left(m_{1}>m_{2}\right)$ are attached to the ends of the string. The system is released for rest. The angular speed of the wheel when $\mathrm{m}_{1}$ descents by a distance h is:

(B) $\left[\frac{2\left(m_{1}-m_{2}\right) g h}{\left(m_{1}+m_{2}\right) R^{2}+1}\right]^{\frac{1}{2}}$

(A) $\left[\frac{2\left(m_{1}+m_{2}\right) g h}{\left(m_{1}+m_{2}\right) R^{2}+1}\right]^{\frac{1}{2}}$

(C) $\left[\frac{\left(m_{1}-m_{2}\right)}{\left(m_{1}+m_{2}\right) R^{2}+1}\right]^{\frac{1}{2}} g h$

(D) $\left[\frac{\left(m_{1}+m_{2}\right)}{\left(m_{1}+m_{2}\right) R^{2}+1}\right]^{\frac{1}{2}} g h$

JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Rotational Mechanics

A uniformly thick wheel with moment of inertia I and radius R is free to rotate about its centre of mass (see fig). A massless string is wrapped over its rim and two blocks of masses

m_{1}

and

m_{2} (m_{1} > m_{2})

are attached to the ends of the string. The system is released for rest. The angular speed of the wheel when

m_{1}

descents by a distance h is:

(A) $[\frac{2 ( m _{1} + m _{2} ) g h}{( m _{1} + m _{2} ) R ^{2} + 1}]^{\frac{1}{2}}$
(B) $[\frac{2 ( m _{1} - m _{2} ) g h}{( m _{1} + m _{2} ) R ^{2} + 1}]^{\frac{1}{2}}$ ✓ Correct
(C) $[\frac{( m _{1} - m _{2} )}{( m _{1} + m _{2} ) R ^{2} + 1}]^{\frac{1}{2}} g h$
(D) $[\frac{( m _{1} + m _{2} )}{( m _{1} + m _{2} ) R ^{2} + 1}]^{\frac{1}{2}} g h$

Answer: (B)

Solution

Let the wheel turn with angular speed

ω

once

m_{1}

has descended a height

h

; since the string does not slip, each block moves with speed

v = ω R

and

m_{2}

rises by the same

h

. Energy is conserved, so the net loss of gravitational potential energy equals the total kinetic energy gained:

(m_{1} - m_{2}) g h = \frac{1}{2} (m_{1} + m_{2}) v^{2} + \frac{1}{2} I ω^{2}

. Substituting

v = ω R

gives

(m_{1} - m_{2}) g h = \frac{1}{2} [(m_{1} + m_{2}) R^{2} + I] ω^{2}

, so

ω = \frac{2 ( m _{1} - m _{2} ) g h}{( m _{1} + m _{2} ) R ^{2} + I}

JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Single correct option

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JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Rotational Mechanics

Solution

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