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A playground merry-go-round has radius 2.40 m and moment of inertia 2100 \[kg{{m}^{2}}\] about a vertical axle through its center, and it turns with negligible friction. A child applies an 18.0-N force tangentially to the edge of the merry-go-round for 15.0 s. find the work done by the child?
A- 200 J
B- 20 J
C- 100 J
D- 400 J

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Last updated date: 14th Apr 2024
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MVSAT 2024
Answer
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Hint: Merry go round always moves in a circular motion, it is fixed. Now the force which is bringing it into circular motion here is the 18 N force acting tangentially and the angle between the force and the radius vector, they are perpendicular to each other. So, this force acts as a torque. We know the relationship between torque, moment of inertia and angular velocity.

Complete Step by step solution:
Force$=18N$
Radius,$r=2.4m$
Moment of inertia, \[I=2100kg{{m}^{2}}\]
Torque,
$\tau =rF \\
\Rightarrow \tau =2.4\times 18 \\
\therefore \tau =43.2Nm \\$
Also, the relationship between torque and moment of inertia and angular acceleration.
$\tau =I\alpha \\
\Rightarrow \alpha =\dfrac{\tau }{I} \\
\Rightarrow \alpha =\dfrac{43.2}{2100} \\
\therefore \alpha =0.021rad{{s}^{-2}} \\$
Now we know the time for which this force acts. For $15s$
$\alpha =\dfrac{\omega }{t} \\
\Rightarrow \omega =\alpha t \\
\Rightarrow \omega =0.021\times 15 \\
\therefore \omega =0.315rad{{s}^{-1}} \\$
Now using work energy theorem change in kinetic energy must be equal to work done. The merry go round starts from rest.
$\Rightarrow W=\Delta K \\
\Rightarrow W={{K}_{f}}-{{K}_{i}} \\
\Rightarrow W={{K}_{f}}-0 \\
\Rightarrow W={{K}_{f}} \\
\Rightarrow W=\dfrac{I{{\omega }^{2}}}{2} \\
\Rightarrow W=\dfrac{2100\times {{0.315}^{2}}}{2} \\
\therefore W=100J \\$

So, the correct option is (c)

Note: While substituting the values all the units must be in the same notation preferably in SI. If the frequency is given in round per minute, always convert it into round per second and if we multiply it by 2 \[\pi \] we get the angular frequency.
Also, torque is a vector quantity and is given by the cross product of force and perpendicular distance from the axis of rotation. In this problem both the vectors were perpendicular to each other, we have to see these all the things.