SATHEE: Rotational Motion Question 38

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

B) g

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

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

Show Answer

Answer:

Correct Answer: D

Solution:

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.

$F = m a$

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

$m R α = m g - T$

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

$I = m R^{2}$

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

$τ = m R^{2} α \Rightarrow τ = m R^{2} (a R) \Rightarrow τ = m R a$

Which is the rotational torque.

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

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

$m R a = T R \Rightarrow m a = T$

Now, the tension in the string is given by

$T = m g - m a$

Therefore, equating both the equations,

we get⇒ $m a = m g - m a \Rightarrow 2 m a = m g \Rightarrow 2 a = g$

$∴ a = g 2$

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

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?

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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?

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu

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?