Question

A 5 kg uniform disk with a 40 cm radius swings without friction about a nail through a point 10 cm from the rim. It is released from rest from a position with its center above the nail.(a) What is its angular velocity when the center is directly below the nail?(b) What is its angular acceleration when the center is level with the nail?(c) Suppose it hangs with its center directly below the nail. If it is pushed a small distance from this equilibrium position then what is its period of small oscillations?

Answer 1

(a) Angular velocity is 8.31 rad/s

(b) Angular acceleration is 17.29 rad/s²

(c) Period of oscillation is 1.51 sec

Explanation

Given:

Mass, m = 5 kg

Radius, r = 40 cm

Distance, x = 10 cm

(a)

Angular velocity, ω = ?

In this case, energy is conserved.

$\frac{1}{2} I w^2 = mg (2a)\\\\w^2 = \frac{4mag}{I} \\\\w = \sqrt{\frac{4mag}{I} }$

where,

I = moment of Inertia

g = gravity

a = distance between the nail and the center

To find I of nail:

$I_n_a_i_l = I_c_e_n_t_e_r + ma^2$

and

$I_c_e_n_t_e_r = \frac{mr^2}{2} \\\\I_c_e_n_t_e_r = \frac{5 X 0.4^2}{2} \\\\I_c_e_n_t_e_r = 0.4 kg.m^2$

On substituting the value:

$I_n_a_i_l = 0.4 + 5 ( 0.4 - 0.1) ^2\\\\I_n_a_i_l = 0.4 + 0.45\\\\I_n_a_i_l = 0.85 kg.m^2$

So, ω is:

$w = \sqrt{\frac{4 X 5X 9.8 X 0.3}{0.85} } \\\\w = 8.31 rad/s$

(b)

Angular acceleration, α = ?

We know:

$\alpha = \frac{t}{I}$

where,

ζ = torque

I = moment of Inertia

To calculate torque:

$t = F X a\\ \\ = mg X a\\\\= 5 X 9.8 X 0.3\\\\= 14.7 Nm$

Substituting the value we get:

$\alpha = \frac{14.7}{0.85} \\\\\alpha = 17.29 rad/s^2$

Thus, the angular acceleration is 17.29 rad/s²

(c)

Period of oscillation, T = ?

We know:

$T = 2\pi \sqrt{\frac{I}{mga} }$

On substituting the value:

$T = 2 X 3.14 X\sqrt{\frac{0.85}{5 X 9.8X0.3} } \\\\T = 1.51 s$