Answer :
(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.
[tex]\frac{1}{2} I w^2 = mg (2a)\\\\w^2 = \frac{4mag}{I} \\\\w = \sqrt{\frac{4mag}{I} }[/tex]
where,
I = moment of Inertia
g = gravity
a = distance between the nail and the center
To find I of nail:
[tex]I_n_a_i_l = I_c_e_n_t_e_r + ma^2[/tex]
and
[tex]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[/tex]
On substituting the value:
[tex]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[/tex]
So, ω is:
[tex]w = \sqrt{\frac{4 X 5X 9.8 X 0.3}{0.85} } \\\\w = 8.31 rad/s[/tex]
(b)
Angular acceleration, α = ?
We know:
[tex]\alpha = \frac{t}{I}[/tex]
where,
ζ = torque
I = moment of Inertia
To calculate torque:
[tex]t = F X a\\ \\ = mg X a\\\\= 5 X 9.8 X 0.3\\\\= 14.7 Nm[/tex]
Substituting the value we get:
[tex]\alpha = \frac{14.7}{0.85} \\\\\alpha = 17.29 rad/s^2[/tex]
Thus, the angular acceleration is 17.29 rad/s²
(c)
Period of oscillation, T = ?
We know:
[tex]T = 2\pi \sqrt{\frac{I}{mga} }[/tex]
On substituting the value:
[tex]T = 2 X 3.14 X\sqrt{\frac{0.85}{5 X 9.8X0.3} } \\\\T = 1.51 s[/tex]