A wheel of radius 8 cm is attached to a support so as to rotate about a horizontal axis through its centre. A string of negligible mass wrapped arounf its circumference carries a mass of 0.4 kg attached to its free end. When the mass is released, it descends through 1m in 10 seconds, then its moment of inertia is (Acceleration due to gravity, `g = 10 ms^(-2)`)

Options

A. `1.277 kg m^(2)`
B. `2.177 kg m^(2)`
C. `21.77 kg m^(2)`
D. `12.77 kg m^(2)`

Answer

A

Explanation

Given, radius of wheel, `R = 8 xx 10^(-2)m`, mass of weight, `m = 0.4 kg`, descending length, L = 1 m and time, t = 10 s
As, torque `tau = F.R = mg R`
putting the given values, we get
`= 0.4 xx 10 xx 8 xx 10^(-2)`
So, `tau = 0.32 Nm`
Let the wheel is rotated by an angle `theta`,
`theta = (a rc)/("radius") = (1)/(0.08) = 12.5` rad
So, from the equation of rotational motion,
`theta = omega_(0) t + (alpha t^(2))/(2)`
`rArr alpha = (2 theta)/(t^(2)) = (2 xx 12.5)/(10 xx 10) = 0.25 "rad s"^(-2)`
`[because omega_(0) = 0]`
Now, as
Torque, `tau = I alpha`
`:.` Moment of inertia of a wheel,
`I = (tau)/(alpha) = (0.32)/(0.25) = 1.28 kg-m^(2)`
Hence, the correct option is (a).