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A uniform solid cylindrical roller of mass “m” is being pulled on a horizontal surface F parallel to the surface and applied at its center. If the acceleration of the cylinder is “a” and is rolling without slipping, then value of “F” is:
(A) ma
(B) $ \dfrac{5}{3} $ ma
(C) 2ma
(D) $ \dfrac{3}{2} $ ma

Last updated date: 29th Feb 2024
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IVSAT 2024
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Hint: We know that frictional force works in a direction opposite to force, so we get
 $ F-f=ma $
Now, for rotational motion, torque is required to produce an angular acceleration of an object. Therefore $ $ Torque $ =\text{I}\alpha $
Torque $ =r\times F $ Here F= linear force
On comparing these equations and substituting the value of I for the cylindrical area, we can find the value of F.

Complete step by step solution
We know that Force F=ma
When frictional force is working in direction opposite to F, then equation becomes
F $ - $ f=ma…….. (1) Where f=frictional force
Now, we know that
Torque $ =I\alpha $ …… (2)
Here I=moment of inertia and $ \alpha $ =angular acceleration
Also, Torque=R $ \times $ f…… (3)
Comparing equations (2) and (3), we get
I $ \alpha $ =R f
For a cylindrical roller
I $ =\dfrac{1}{2}m{{R}^{2}} $
 $ \dfrac{1}{2}m{{R}^{2}}\alpha = $ R f
 $ \dfrac{1}{2}mR\alpha = $ f……. (4)
Putting value of equation (4) in equation (1), we get
F $ -\dfrac{1}{2}mR\alpha =ma $ ……. (5)
As the cylinder is rolling without slipping the acceleration of point of contact will be zero.
 $ \alpha =\dfrac{a}{R} $
Putting this value in equation (5), we get
F $ -\dfrac{1}{2}mR\alpha =ma $
F= $ ma+\dfrac{1}{2}ma $
 $ =\dfrac{3}{2}ma $ .

Rolling without slipping is a combination of translational and rotational motion where the point of contact is instantaneously at rest.
When an object experiences pure translational motion, all of its points move with the same velocity as the center of mass, that is, in the same direction and with the same speed.
V(r)=v(center of mass)
The object will also move in a straight line in the absence of net external force.
When the object experiences pure rotational motion about its center of mass, all the points move at right angles to the radius in a plane perpendicular to the axis of rotation with speed proportional to distance from the axis of rotation.
V(r)= r w
Thus points on opposite sides of the axis move in opposite directions, points on the axis do not move at all since r=0
V (center of mass)=0
And points on the outer edge move at maximum speed.
When object experiences rolling motion, the point of the object in contact with surface instantaneously at rest
V (point of contact)=0
So, the center of mass of the object moves with speed v (center of mass) =R w and the point farthest from the point of contact moves with twice the speed v (opposite the point of contact) =2R w.

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