Torque and Angular Momentum Physics System of Particles and Rotational Motion Physics Class 11th

What is torque and how does it relate to angular momentum?

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Answered by

Syed Aquib Ur Rahman | Contributor-Level 10

2 months ago

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

$? = \frac{d L}{d t}$

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The dimensions of angular impulse is equal to

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Angular impulse = Change in angular momentum

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By conservation of Angular momentum

Li = Lf

$M R^{2} ω = (m R^{2} + 2 m R^{2}) ω'$

$\frac{2 M}{M + 2 m} = ω'$

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A triangular plate is shown. A force F = 4i - 3j is applied at point P. The torque at point P with respect to point 'O' and 'Q' are :

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

Force vector (F) = 4i - 3j

Torque about O (τ_o) = r_o × F
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Position vector about Q (r_q) = -5i + 5√3j

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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 ]

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Kindly consider the following Image

A particle of mass 'm' is moving in time 't' on a trajectory given by r? = 10αt² i? + 5β(t-5) j?. Where α and β are dimensional constants. The angular momentum of the particle become the same as it was for t = 0 at time t = ______ seconds.

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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
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⇒ 50t (10 - t) = 0
⇒ t = 10 second

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