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Question: The angular momentum of a particle about origin is varying as L = 4t + 8 (SI units) when it moves al...

The angular momentum of a particle about origin is varying as L = 4t + 8 (SI units) when it moves along a straight line y = x - 4 (x, y in meters). The magnitude of force acting on the particle would be:
A. 1 N
B. 2 N
C. 2\sqrt{2} N
D. 3\sqrt{3} N

Explanation

Solution

We are given the angular momentum L and the rate of change of angular momentum should give us torque. Find the way of converting the given direction of its motion into a vector then use the torque formula to find the force.

Formula used:
The rate of change of angular momentum is torque:
τ=dLdt\tau = \dfrac{dL}{dt}.
Torque acting on a body is also given by relation:
τ=r×F\vec{\tau} = \vec{r} \times \vec{F}.

Complete answer:
For the given particle, we have the angular momentum L = 4t + 8. This means that it is changing with time because as the value of t changes, L changes too. Therefore, the rate of change of angular momentum becomes:
τ=dLdt=d(4t+8)dt=4\tau = \dfrac{dL}{dt} = \dfrac{d(4t + 8)}{dt} = 4.
This means that the magnitude of torque is 4 units.

Now, the particle happens to be moving along a straight line y = x -4. This means that the position of the particle at any point is given by this equation. So, we can find the vector from the origin in the following manner:
1. Finding any two points that satisfy the equation for this line along which the particle moves. If we keep x = 0 we get y = - 4. If we keep x = 2, we get y = -2.
2. Now, our required vector will be joining these two points so, the vector going from (0, -4) to (2, -2) can be written as:
r=(20)i^+(2(4))j^=2i^+2j^\vec{r} = (2-0) \hat{i} + (-2 - (-4)) \hat{j} = 2 \hat{i} + 2 \hat{j} .
The magnitude of this vector can be written as:
r=22+22=22r = \sqrt{2^2 +2^2} = 2\sqrt{2}.

The magnitude of torque can be written simply as:
τ=rF\tau = rF,
where F is the magnitude of force acting and r is the perpendicular distance from the axis.
So, we may write:
F=τr=422=2F = \dfrac{\tau}{r} = \dfrac{4}{2\sqrt{2}} = \sqrt{2}N.

So, the correct answer is “Option C”.

Note:
The tricky part in the question is to find out the vector vecrvec{r}, which is supposed to be the distance perpendicular to which the force is acting. Also, as the question says 'angular momentum is varying as L = 4t + 8', one might assume that this is the rate of change of angular momentum but clearly as it is written in terms of L it is the expression for angular momentum itself.