A block A of mass 5 kg hangs from a…

A block A of mass 5 kg hangs from a light string that passes over a uniform disc pulley of mass 4 kg and radius 0.4 m (free to rotate about its fixed…

Numerical answer

A block A of mass

5\ \mathrm{kg}

hangs from a light string that passes over a uniform disc pulley of mass

4\ \mathrm{kg}

and radius

0.4\ \mathrm{m}

(free to rotate about its fixed horizontal axis, string does not slip on it). The other end is tied to a block B of mass

3\ \mathrm{kg}

lying on a rough incline of angle

37^\circ

(

\sin37^\circ=0.6

) with coefficient of kinetic friction

0.5

; the string is parallel to the incline. The system is released from rest and A descends. Take

g=10\ \mathrm{m\,s^{-2}}

. The angular acceleration of the pulley (in

\mathrm{rad\,s^{-2}}

) is ______.

Show answer and solution

Answer: 5

Treat A, B and the pulley separately. For B moving up the incline:

T_2-m_2g\sin\theta-\mu m_2g\cos\theta=m_2a

, where

m_2g\sin\theta=18\ \mathrm N

and friction

=0.5\times3\times10\times0.8=12\ \mathrm N

. For A:

m_1g-T_1=m_1a

. For the pulley (no slipping):

(T_1-T_2)R=\tfrac12MR^2\cdot\dfrac aR

, i.e.

T_1-T_2=\tfrac12Ma

. Adding the three equations eliminates the tensions:

m_1g-18-12=\left(m_1+m_2+\tfrac M2\right)a

, giving

a=\dfrac{50-30}{5+3+2}=2\ \mathrm{m\,s^{-2}}

. Check that B does move up: the net driving force

50\ \mathrm N

exceeds the resisting

30\ \mathrm N

. The angular acceleration of the pulley is

\alpha=a/R=2/0.4=5\ \mathrm{rad\,s^{-2}}

. Tensions:

T_1=40\ \mathrm N

T_2=36\ \mathrm N

, difference

4\ \mathrm N=\tfrac12Ma

, consistent.

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Original question written in the JEE Advanced 2022 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake