A particle of mass '$m$' is moving in time '$t$' on a trajectory given by $$\vec{r}=10 \alpha t^{2} \hat{i}+5 \beta(t-5) \hat{j}$$ Where $\alpha$ and $\beta$ are dimensional constants. The angular momentum of the particle become the same as it was for $t=0$ at time t = ____ seconds.

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Rotational Mechanics

A particle of mass '

m

' is moving in time '

t

' on a trajectory given by

r = 10 α t^{2} \hat{i} + 5 β (t - 5) \hat{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.

Answer: 10

Solution

Differentiating the position,

v = 20 α t \hat{i} + 5 β \hat{j}

. The motion is confined to the

x y

plane, so the angular momentum about the origin has only a

z

component:

L_{z} = m (x v_{y} - y v_{x}) = m [10 α t^{2} (5 β) - 5 β (t - 5) (20 α t)] = 50 m α β (10 t - t^{2})

t = 0

this is zero, and it returns to zero when

t (10 - t) = 0

. The non-zero root gives

t = 10

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Integer answer

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