Torque and Angular Momentum

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

Kindly consider the following Image

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. 

Both torque and angular momentum have direction. We determine their direction based on the right-hand rule. That makes them vector quantities in rotational motion. 

Yes, an object can have angular momentum without torque. The physics behind it is that if no external force act on an object, its angular momentum is constant. That is based on the law of conservation of momentum. 

Torque is the measure of rotational force on an object (rigid or extended body). It changes the object's angular momentum. To calculate in physics, torque equals the rate of change of angular momentum. The formula for that is

$?=\frac{dL}{dt}$

r = 1.5 m

F = 12t – 3t2 N

l = 4.5 kgm²

$\alpha =\frac{T}{l}=\frac{\frac{2}{3}\left(12t-3t^2\right)}{4.5}=\frac{1}{3}\left(12t-3t^2\right)$

$\Rightarrow \alpha =4t-t^2$

$\Rightarrow \frac{dw}{dt}=ut-t^2={\displaystyle \underset{0}{\overset{0}{\int }}dw={\displaystyle \int {\left(4t-t^2\right)}_0^{t}dt}}$

No. of rev = $\frac{36}{2\pi }=\frac{18}{\pi }=\frac{k}{\pi }\Rightarrow k=18$