Consider a 20 kg uniform circular disk of radius 0.2 m. It is pin supported at its centre and is at rest initially. The disk is acted upon by a constant force F = 20 N through a massless string wrapped around its periphery as shown in the figure. Suppose the disk makes n number of revolutions to attain an angular speed of 50 rad s-1. The value of n, to the nearest integer, is ---------.

[Given : In one complete revolution, the disk rotates by 6.28 rad ]

 

Angular impulse = Change in angular momentum

[J] = [mvr]

[J] = [M¹L²T^–1]

By conservation of Angular momentum

Li = Lf 

$M{R}^2\omega =\left(m{R}^2+2m{R}^2\right)\omega \text{'}$

$\frac{2M}{M+2m}=\omega \text{'}$

$\Rightarrow$ $\omega \text{'}=\frac{2M}{M+2m}$

Position vector about O (r_o) = 5i + 5√3j

Force vector (F) = 4i - 3j

Torque about O (τ_o) = r_o × F
τ_o = (5i + 5√3j) × (4i - 3j)
τ_o = -15k - 20√3k = (-15 - 20√3)k

Position vector about Q (r_q) = -5i + 5√3j

Torque about Q (τ_q) = r_q × F
τ_q = (-5i + 5√3j) × (4i - 3j)
τ_q = 15k - 20√3k = (15 - 20√3)k

r? = 10αt²î + 5β (t-5)?
v? = dr? /dt = 20αtî + 5β?
As L? = m (r? × v? )
So, at t=0, L=0
given L is same at t=t as at t=0
⇒ r? × v? = 0
⇒ (10αt²î + 5β (t-5)? ) × (20αtî + 5β? ) = 0
⇒ 50αβt² (î×? ) + 100αβt (t-5) (? ×î) = 0
⇒ 50αβt² k? - 100αβt (t-5) k? = 0
⇒ 50t² - 100t (t-5) = 0
⇒ 50t² - 100t² + 500t = 0
⇒ -50t² + 500t = 0
⇒ 50t (10 - t) = 0
⇒ t = 10 second

The direction of torque and angular momentum defines how and in which orientation an object will rotate or sustain its spin. This is important to understand in machines, athletic movements, and even natural phenomena, such as planetary motion.