Answer
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Hint: 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:
$\tau = \dfrac{dL}{dt}$.
Torque acting on a body is also given by relation:
$\vec{\tau} = \vec{r} \times \vec{F}$.
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:
$\tau = \dfrac{dL}{dt} = \dfrac{d(4t + 8)}{dt} = 4$.
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:
$\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:
$r = \sqrt{2^2 +2^2} = 2\sqrt{2}$.
The magnitude of torque can be written simply as:
$\tau = rF$,
where F is the magnitude of force acting and r is the perpendicular distance from the axis.
So, we may write:
$F = \dfrac{\tau}{r} = \dfrac{4}{2\sqrt{2}} = \sqrt{2}$N.
So, the correct answer is “Option C”.
Note:
The tricky part in the question is to find out the vector $vec{r}$, 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.
Formula used:
The rate of change of angular momentum is torque:
$\tau = \dfrac{dL}{dt}$.
Torque acting on a body is also given by relation:
$\vec{\tau} = \vec{r} \times \vec{F}$.
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:
$\tau = \dfrac{dL}{dt} = \dfrac{d(4t + 8)}{dt} = 4$.
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:
$\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:
$r = \sqrt{2^2 +2^2} = 2\sqrt{2}$.
The magnitude of torque can be written simply as:
$\tau = rF$,
where F is the magnitude of force acting and r is the perpendicular distance from the axis.
So, we may write:
$F = \dfrac{\tau}{r} = \dfrac{4}{2\sqrt{2}} = \sqrt{2}$N.
So, the correct answer is “Option C”.
Note:
The tricky part in the question is to find out the vector $vec{r}$, 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.
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