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} $
Show Answer
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 .
BETA
