A wheel of radius `6 cm` is mounted so as to rotate about a horizontal axis through its centre. A string of negligibe mass wrapped round its circumference carries a mass of `0.2 kg` attrached to its free end. When let fall, the mass descends through `1 m` in `5 s`. Calculate the angular acceleration of the wheel, its moment of inertia and tension in the cord.

Explanation

Here, `r = 6 cm = 6 xx 10^(-2) m`
`m = 0.2kg, s = 1m, t = 5s`.
If `a` is acceleration of falling mass, Fig.
from `s = ut + (1)/(2)at^(2)`
`1 = 0+ (1)/(2) xx a xx 5^(2), a = (2)/(25)m//s^(2)`
Angular acc. `alpha = (a)/(r ) = (2//25)/(6 xx 10^(-2)) = 1.33 rad//s^(2)`
Velocity after `5 s, v = u + at`
`= 0 + (2)/(25) xx 5 = 0.4 m//s`
Angular velocity `omega = (v)/(r ) = (0.4)/(6 xx 10^(-2)) = (20)/(3) rad//s`
According to the law of conservation of energy
Loss of `P.E =` Increase in (K.E. of translation `+ K.E.` of rotation)
`mgh = (1)/(2)mv^(2) + (1)/(2)I omega^(2)`
`0.2 xx 9.8 xx 1 = (1)/(2) xx 0.2 (0.4)^(2) + (1)/(2)I ((20)/(3))^(2)`
`0.196 = 0.1 xx 0.16 + I xx (400)/(18)`
`0.180 = I xx (400)/(18)`
or `I = (0.180 xx 18)/(400) = 8.75 xx 10^(-2) lg m^(2)`
If `I` is tension in the string, then from `ma = mg - T`
`T = mg - ma = m(g -a)`
`= 0.2 (9.8 - 0.08) = 0.2 xx 9.72`
`T = 1.944N`