Question

A solid cyinder of mass `m=4 kg` and radius `R=10 cm` has two ropes wrapped around it. one near each end. The cylinder is held horizontally by fixing the two free ends of the cords to the hooks on the ceiling such that both the cords are exactly vertical. The cylinder is released to fall under gravity. Find the linear acceleration of the cylinder.

Answer 1

Method 1: Force/Torque Method:
Let `a=` linear acceleration and `alpha=` angular acceleration of the cylinder. For the linear motion of the cylinder.
`mg-2T=ma`
For the rotational motion: Net torque `=Ialpha`

Also, the linear acceleration of the cyinder is the same as the tangential acceleration of an point its surface, `a=Ralpha`
Combining the three equations, we get
`mg=ma+m/2aimpliesa=2/3g`
Method 2: Energy method
The motion of the cylinder is rotational and translation combined . Using conservation of mechanical energy
`/_K=/_U=0`

`/_K+/_U=0`
or Loss in `PE =` gain in `KE`
`mgh=1/2mv^(2)+1/2Iomega^(2)`
Constraint relation
Velocity of point `P v_(P)=0v-omegaR`
`omega=v/R` .........i
From eqn and i and ii `mgh=1/2mv^(2)+1/2((mR)/2(v/R)^(2)`
`=m/2 (v^(2)+(v^(2))/2)`
`2gh=3/2v^(2)implies v^(2)=4/3gh` ..... ii
Differentiating eqn iii with respect to time, we get
`2v(dv)/(dt)=4/3g(dh)/(dg)`
`22va=4/3gvimpliesa=2/3g`