Question

A uniform solid cylinder is at rest on a horizontal surface. A small particle of mass m (which is connected to massless string) is held in
equilibrium by applying the horizontal force F as shown in figure. The value of force F will be (Take acceleration due to gravity to be
g)
F
m
(A)
mg
2
(B) mg
(C)
3mg​

Answer 1

Explanation:

ANSWER

Let the mass of cylinder is m and radius of it is r.

so, moment of inertia of cylinder , I=

2

1

mr

2

torque = moment of inertia * angular acceleration

force * displacement =

2

1

mr

2

×

r

a

we know, linear acceleration, a = r * angular acceleration

T×r=

2

1

mr×a

T=

2

1

ma ......(1)

now, net force = ma

weight - tension = ma

or, mg−

2

1

ma=ma

or, mg=

2

3

ma

or, a=

3

2g

hence, tension in the string =

2

1

ma=

3

mg

and acceleration =

3

2g