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.
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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1 Answer
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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
⇒ 50t² - 100t² + 500t = 0
⇒ -50t² + 500t = 0
⇒ 50t (10 - t) = 0
⇒ t = 10 second
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Kindly consider the following Image
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.
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