Rotational Motion Question 38

Question: A mass ‘m’ is 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?

[JEE MAIN 2014]

Options:

A) $ \frac{5g}{6} $

B) g

C) $ \frac{2g}{3} $

D) $ \frac{g}{2} $

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Answer:

Correct Answer: D

Solution:

  • [d]

A mass ‘m’ is attached by a massless string that is wounded around a uniform hollow cylinder.

Now, according to Newton’s second law of motion, the force acting on a body is equal to the product of mass and acceleration.

$F=ma$

Now, if the tension is produced in the string, than the force is given byF=mg−TNow, putting the value of F in the above equation, we getma=mg−TNow, in the case, the string is not slipping on the cylinder.

Therefore, the freefall of the suspended mass will be due to acceleration due to gravity.

The linear acceleration a will rise tangentially with the angular acceleration of the string, therefore, the acceleration is given bya=RαPutting this value of acceleration in the above equation, we get

$mRα=mg−T$

Now, the torque produced by the angular movement of the string is the result of the angular acceleration of the cylinder.

Therefore, the torque is given byτ=I×αHere, I is the rotational inertia in the cylinder and is given by

$I=mR^2$

Putting this value, the equation of the torque is given by

$τ=mR^2α⇒τ=mR^2(aR)⇒τ=mRa$

Which is the rotational torque.

This torque also produces a linear torque and is given by $τ=TR$

Now, equating the equations of $τ$ , we get

$mRa=TR⇒ma=T$

Now, the tension in the string is given by

$T=mg−ma$

Therefore, equating both the equations,

we get⇒$ma=mg−ma⇒2ma=mg⇒2a=g$

$∴a=g2$

Therefore, the mass suspended will fall on release with the acceleration g2 .



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