A torque of 20Nm is applied on a wheel having angular momentum of $2\dfrac{{Kg{m^2}}}{s}$, calculate the angular momentum of the wheel after 4 seconds.
$\left( A \right)$ 80 $\dfrac{{Kg{m^2}}}{s}$
$\left( B \right)$ 55 $\dfrac{{Kg{m^2}}}{s}$
$\left( C \right)$ 45 $\dfrac{{Kg{m^2}}}{s}$
$\left( D \right)$ 22 $\dfrac{{Kg{m^2}}}{s}$
Answer
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Hint – In this question use the concept that torque is the rate of change of angular momentum with respect to the time and the rate of change of angular momentum is the difference of the final momentum to the initial momentum so let the final momentum of the wheel be x$\dfrac{{Kg{m^2}}}{s}$. This will help approaching the problem.
Complete step-by-step answer:
Given data:
Torque = 20 Nm
Angular momentum = 2 $\dfrac{{Kg{m^2}}}{s}$.
Now we have to calculate the angular momentum after 4 seconds.
Now we all know torque is the rate of change of angular momentum with respect to the time.
Rate of change of angular momentum is the difference of the final momentum to the initial momentum.
Now let the final momentum of the wheel be x$\dfrac{{Kg{m^2}}}{s}$.
So the rate of change of angular momentum is equal to = final angular momentum – initial angular momentum.
Now substitute the values we have,
The rate of change of angular momentum = x$\dfrac{{Kg{m^2}}}{s}$– 2$\dfrac{{Kg{m^2}}}{s}$.
Now the given time interval is 4 sec.
Therefore, 20 Nm = [x$\dfrac{{Kg{m^2}}}{s}$– 2$\dfrac{{Kg{m^2}}}{s}$] / 4
Now simplify this we have,
$ \Rightarrow 20 \times 4 = \left( {x - 2} \right)$ $Kgm^2/s$
Now substitute (-2) to the L.H.S and simplify we have
Therefore, x = (80 + 2) $\dfrac{{Kg{m^2}}}{s}$
Therefore, x = 82 $\dfrac{{Kg{m^2}}}{s}$.
So this is the required angular momentum of the wheel after 4 seconds.
So this is the required answer.
Hence option (D) is the correct answer.
Note – Torque can simply be considered as a twisting force that causes rotation. Couple force can be considered as a subset of torque force only as this equal and opposite force too causes rotation too. When a particle is rotating in a circular direction then it tends to rotate with some acceleration this acceleration is called the angular acceleration.
Complete step-by-step answer:
Given data:
Torque = 20 Nm
Angular momentum = 2 $\dfrac{{Kg{m^2}}}{s}$.
Now we have to calculate the angular momentum after 4 seconds.
Now we all know torque is the rate of change of angular momentum with respect to the time.
Rate of change of angular momentum is the difference of the final momentum to the initial momentum.
Now let the final momentum of the wheel be x$\dfrac{{Kg{m^2}}}{s}$.
So the rate of change of angular momentum is equal to = final angular momentum – initial angular momentum.
Now substitute the values we have,
The rate of change of angular momentum = x$\dfrac{{Kg{m^2}}}{s}$– 2$\dfrac{{Kg{m^2}}}{s}$.
Now the given time interval is 4 sec.
Therefore, 20 Nm = [x$\dfrac{{Kg{m^2}}}{s}$– 2$\dfrac{{Kg{m^2}}}{s}$] / 4
Now simplify this we have,
$ \Rightarrow 20 \times 4 = \left( {x - 2} \right)$ $Kgm^2/s$
Now substitute (-2) to the L.H.S and simplify we have
Therefore, x = (80 + 2) $\dfrac{{Kg{m^2}}}{s}$
Therefore, x = 82 $\dfrac{{Kg{m^2}}}{s}$.
So this is the required angular momentum of the wheel after 4 seconds.
So this is the required answer.
Hence option (D) is the correct answer.
Note – Torque can simply be considered as a twisting force that causes rotation. Couple force can be considered as a subset of torque force only as this equal and opposite force too causes rotation too. When a particle is rotating in a circular direction then it tends to rotate with some acceleration this acceleration is called the angular acceleration.
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