A block X of mass 0.5 kg is held by a long massless string on a frictionless inclined plane of inclination `30^@` to the horizontal. The string is wound on a uniform solid cylindrical drum Y of mass 2kg and of radius 0.2 and of radius 0.2 m as shown in Fingure. The drum is given an initial angular velocity such that the block X starts moving up the plane.
(i) Find the tension in the string during the motion.
(ii) At a certain instant of time the magnitude of the angular velocity of Y s `10 rad s^(-1)` calculate the distance travelled by X from that instant of time until it comes to rest

Answer

`1.63 N, 1.224 m`

Explanation

(a) Let a be the linear retardation of block `X` and `alpha` is the angular retardation of drum `Y`. Then, `alpha` is the angular retardation of drum `Y`. Then,
`a = R alpha`
`mg sin 30^@ - T = ma` ….(1)
`(mg)/(2) - T = ma` ….(2)
`alpha = (tau)/((1)/(2) MR^(2))` or `alpha = (2T)/(MR)`...(3)
solving Eqs. (1),(2) and (3) for `T`, we get
`T = (1)/(2) (M mg)/((M + 2m))`
substituting the value, we get

`T = ((1)/(2)) {((2)(0.5)(9.8))/(2 + (0.5) (2))}`
`T = 1.63 N`
(ii) From Equation (3), angular retardation of drum
`alpha =(2T)/(MR) =((2)(1.63))/((2)(0.2)) = 8.15 rad//s^(2)`
or linear retardation of block
`a = R alpha = (0.2) (8.15) = 1.63 m//s^(2)`
At the moment when angular velocity of drum is
`Omega_(0) = 10 rad//s^(2)`
The linear velocity of block will be
`v_(0) = omega_(0) R = (10) (0.2) = 2 m//s`
Now, the distance (s) travelled by the block until it comes to rest will be given by
`s - (v_(0)^(2)) [Using v^(2) = v_(0)^(2) - 2as with v = 0]`
`s =((2)^(2))/(2(1.63)) m`
or `s = 1.22 m`.