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

Why are torque and angular momentum considered vectors?

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Answered by

Syed Aquib Ur Rahman | Contributor-Level 10

2 months ago

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.

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The dimensions of angular impulse is equal to

alok kumar singh

Angular impulse = Change in angular momentum

[J] = [mvr]

[J] = [M¹L²T^–1]

A thin circular ring of mass M and radius R is rotating with a constant angular velocity 2 rads^-1 in a horizontal plane about an axis vertical to its plane and passing through the centre of the ring. If two objects each of mass m be attached gently to the opposite ends of a diameter of ring, the will then rotate with an angular velocity (in rads^-¹).

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By conservation of Angular momentum

Li = Lf

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$\frac{2 M}{M + 2 m} = ω'$

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

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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Force vector (F) = 4i - 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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