
A uniform disc is spun with an angular velocity $\vec \omega $ and simultaneously projected with a linear velocity $v$ towards left on a plank, while the plank moves towards right with a constant velocity $2v$. If the disc rolls without sliding on the plank just after its spinning, find the magnitude of $\vec \omega $ in $\dfrac{{rad}}{{\sec }}.$(Take $v = 3\,m\,{s^{ - 1}},\,R = 1\,m$)
Answer
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Hint: Use the relation between the linear velocity of the object, angular velocity of the object and radius of the circular path of the object. First determine the linear velocity of the disc with respect to the plank. Then determine the magnitude of the angular velocity of the disc using the formula for linear velocity in terms of angular velocity.
Formula used:
The linear velocity of an object is given by
$v = \omega R$ …… (1)
Here, $v$ is the linear velocity of the object, $\omega $ is the angular velocity of the object and $R$ is the radius of the circular path of the object.
Complete step by step answer:
We have given that a uniform disc is spun with an angular velocity $\vec \omega $ and simultaneously projected with a linear velocity $v$ towards left on a plank, while the plank moves towards right with a constant velocity $2v$.
The diagram representing the motion of the disc and the plank is as follows:
We can determine the velocity of the disc with respect to the plank.
Hence, the velocity of the disc is ${v_d} = v + 2v = 3v$
Rewrite equation (1) for the linear velocity of the disc with respect to plank.
${v_d} = \omega R$
Rearrange the above equation for the angular velocity $\omega $ of the disc.
$\omega = \dfrac{{{v_d}}}{R}$
Substitute $3v$ for ${v_d}$ in the above equation.
$\omega = \dfrac{{3v}}{R}$
Substitute $3\,\,m\,{s^{ - 1}}$ for $v$ and $1\,m$ for $R$ in the above equation.
\[\omega = \dfrac{{3\left( {3\,\,m\,{s^{ - 1}}} \right)}}{{1\,m}}\]
\[ \Rightarrow \omega = 9\,{\text{rad}} \cdot {{\text{s}}^{ - 1}}\]
Hence, the magnitude of the angular velocity of the disc is \[9\,{\text{rad}} \cdot {{\text{s}}^{ - 1}}\].
Note:
Since the disc is moving towards the left with a linear velocity, the angular velocity of the disc should also be in the left direction. As the direction of angular velocity of the disc is in the left direction, the angular velocity of the disc will have a negative sign as the left direction represents the negative X-axis. But we have asked the magnitude of the angular velocity, hence, the negative sign is not considered in the solution.
Formula used:
The linear velocity of an object is given by
$v = \omega R$ …… (1)
Here, $v$ is the linear velocity of the object, $\omega $ is the angular velocity of the object and $R$ is the radius of the circular path of the object.
Complete step by step answer:
We have given that a uniform disc is spun with an angular velocity $\vec \omega $ and simultaneously projected with a linear velocity $v$ towards left on a plank, while the plank moves towards right with a constant velocity $2v$.
The diagram representing the motion of the disc and the plank is as follows:
We can determine the velocity of the disc with respect to the plank.
Hence, the velocity of the disc is ${v_d} = v + 2v = 3v$
Rewrite equation (1) for the linear velocity of the disc with respect to plank.
${v_d} = \omega R$
Rearrange the above equation for the angular velocity $\omega $ of the disc.
$\omega = \dfrac{{{v_d}}}{R}$
Substitute $3v$ for ${v_d}$ in the above equation.
$\omega = \dfrac{{3v}}{R}$
Substitute $3\,\,m\,{s^{ - 1}}$ for $v$ and $1\,m$ for $R$ in the above equation.
\[\omega = \dfrac{{3\left( {3\,\,m\,{s^{ - 1}}} \right)}}{{1\,m}}\]
\[ \Rightarrow \omega = 9\,{\text{rad}} \cdot {{\text{s}}^{ - 1}}\]
Hence, the magnitude of the angular velocity of the disc is \[9\,{\text{rad}} \cdot {{\text{s}}^{ - 1}}\].
Note:
Since the disc is moving towards the left with a linear velocity, the angular velocity of the disc should also be in the left direction. As the direction of angular velocity of the disc is in the left direction, the angular velocity of the disc will have a negative sign as the left direction represents the negative X-axis. But we have asked the magnitude of the angular velocity, hence, the negative sign is not considered in the solution.
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