Question 603: Centre of Mass and Rotational Motion

Question

If there is a change of angular momentum from $1 \mathrm{j}$-$\mathrm{s}$ to $4 \mathrm{j}$-$\mathrm{s}$ in $4 \mathrm{~s}$, then the torque

Accepted Answer

The torque ($	au$) can be found by using the relation between torque and the rate of change of angular momentum. This relationship is given by:
$ 	au = \frac{\Delta L}{\Delta t} $
where:
$\Delta L$ is the change in angular momentum,
$\Delta t$ is the change in time.
In this case, the change in angular momentum ($\Delta L$) from $1 \mathrm{j} \, \mathrm{s}$ to $4 \mathrm{j} \, \mathrm{s}$ is :
$ \Delta L = 4 \, \mathrm{j} \, \mathrm{s} - 1 \, \mathrm{j} \, \mathrm{s} = 3 \, \mathrm{j} \, \mathrm{s} $
And the change in time ($\Delta t$) is $4 \, \mathrm{s}$.
Therefore, the torque can be calculated as:
$ 	au = \frac{3 \, \mathrm{j} \, \mathrm{s}}{4 \, \mathrm{s}} = \frac{3}{4} \, \mathrm{J} $
Thus, the correct answer is
Option B
, $\left(\frac{3}{4}ight) \, \mathrm{J}$.

Answer

$\left(\frac{5}{4}\right) \mathrm{J}$

Answer

$\left(\frac{3}{4}\right) \mathrm{J}$

Answer

$1 \mathrm{~J}$

Answer

$\left(\frac{4}{3}\right) \mathrm{J}$

Centre of Mass and Rotational Motion

If there is a change of angular momentum from 1 \mathrm{j}-\mathrm{s} to 4 \mathrm{j}-\mathrm{s} in 4 \mathrm{~s}, then the torque

Choose the correct answer: