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A heavy homogeneous cylinder has mass m ...


A heavy homogeneous cylinder has mass `m` and radius `R`. It is accelerated by a force `F` which is applied through a rope wound around a light drum of radius r attached to the cylinder (figure) the coefficient of static friction is sufficient for the cylinder to roll without slipping.

A

acceleration of centre of cylinder is `a = (2F)/(2mR) (R+r)`

B

friction assumed to be in the opposite direction of F is `f =(2)/(3)((1)/(2)-(r)/(R))F`

C

If `((r)/(R)) lt (1)/(2)` a will be greater than `(F)/(m)`

D

Friction assumed to be in the direction of F is `f = (2)/(3) ((1)/(2) -(r)/(R))F`

Text Solution

Verified by Experts

The correct Answer is:
A, B

`F=f=ma Fr+fR=(mR^(2))/(2)alpha" "a=Ralpha`
Solving `a=(2F)/(2mR) (R+r)f=(2)/(3)((1)/(2)-(r)/(R))F`
If `((r)/(R))gt (1)/(2)`
f will be forward
`r F + f = ma" "a gt (F)/(m)`
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