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A uniform solid cylindrical roller of mass ‘m’ is being pulled on a horizontal surface with force F parallel to the surface and applied at its center. If the acceleration of the cylinder is ‘a’ and it is rolling without slipping then the value of ‘F’ is:
  • a)
    ma
  • b)
    5/3 ma
  • c)
    3/2 ma
  • d)
    2 ma
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
A uniform solid cylindrical roller of mass ‘m’ is being pulled on a h...
From figure, ma = F – f ....(i)
And, torque τ = lα 2
mR2/2 α = fR
Put this value in equation (i),
mα = F - mα/2 or F = 3mα/2
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Most Upvoted Answer
A uniform solid cylindrical roller of mass ‘m’ is being pulled on a h...
Solution:

Given: Mass of cylinder (m), acceleration of cylinder (a), force pulling the cylinder (F)

We know that when a cylinder rolls without slipping, the force of friction acting on it is given by:

f = (1/2) * m * R * a

where R is the radius of the cylinder.

Also, we know that the force pulling the cylinder is given by:

F = (1/2) * m * a + f

Substituting the value of f in the above equation, we get:

F = (1/2) * m * a + (1/2) * m * R * a

= (1/2) * m * (a + R * a)

= (3/2) * m * a

Therefore, the value of F is (3/2) * m * a, which is option C.

Hence, the correct answer is option 'C'.
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A uniform solid cylindrical roller of mass ‘m’ is being pulled on a horizontal surface with force F parallel to the surface and applied at its center. If the acceleration of the cylinder is ‘a’ and it is rolling without slipping then the value of ‘F’ is:a)mab)5/3 mac)3/2 mad)2 maCorrect answer is option 'C'. Can you explain this answer?
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