Question

dad is pushing his son on the merry - go - round in the park. the merry - go - round has a mass of 160 kg and a diameter of 2.40 m; dynamically, it can be treated like a disc. his 35 kg son is sitting 0.5 m from the center of the merry - go - round. if dad pushes with a tangential force of 55 n for 1.4 seconds three times, then how fast is his son moving (tangentially) if the merry - go - round starts from rest? hint: use principles of torque and angular acceleration to answer this question. hint: find angular acceleration first, then find final angular velocity. your answer:

Explanation:

Step1: Calculate the moment of inertia of the system

The merry - go - round is a disc, so its moment of inertia \(I_{mg}= \frac{1}{2}MR^{2}\), where \(M = 160\space kg\) and \(R=\frac{2.40}{2}=1.20\space m\). The son can be treated as a point mass, so his moment of inertia \(I_{son}=mr^{2}\), where \(m = 35\space kg\) and \(r = 0.5\space m\).

First, calculate \(I_{mg}\):
\(I_{mg}=\frac{1}{2}\times160\times(1.20)^{2}=96\times1.44 = 138.24\space kg\cdot m^{2}\)

Then, calculate \(I_{son}\):
\(I_{son}=35\times(0.5)^{2}=35\times0.25 = 8.75\space kg\cdot m^{2}\)

The total moment of inertia \(I = I_{mg}+I_{son}=138.24 + 8.75=146.99\space kg\cdot m^{2}\approx147\space kg\cdot m^{2}\)

Step2: Calculate the torque

The torque \(\tau = F\times R\) (since the force is tangential, the lever arm is the radius of the merry - go - round), where \(F = 55\space N\) and \(R = 1.20\space m\).
\(\tau=55\times1.20 = 66\space N\cdot m\)

Step3: Calculate the angular acceleration

Using the formula \(\tau=I\alpha\), we can solve for angular acceleration \(\alpha\). Rearranging the formula gives \(\alpha=\frac{\tau}{I}\)
\(\alpha=\frac{66}{147}\approx0.449\space rad/s^{2}\)

Step4: Calculate the total time of application of force

The force is applied for \(t = 1.4\space s\) three times, so the total time \(T=3\times1.4 = 4.2\space s\)

Step5: Calculate the final angular velocity

Using the formula \(\omega=\omega_{0}+\alpha T\), where \(\omega_{0} = 0\) (starts from rest)
\(\omega=0 + 0.449\times4.2\approx1.886\space rad/s\)

Step6: Calculate the tangential speed of the son

The tangential speed \(v=\omega r\), where \(r = 0.5\space m\) (the distance of the son from the center)
\(v=1.886\times0.5\approx0.943\space m/s\)

Wait, let's check the moment of inertia calculation again. Wait, the radius for the torque: the force is applied tangentially to the merry - go - round, so the lever arm is the radius of the merry - go - round (\(R = 1.2\space m\)). But when calculating the moment of inertia of the merry - go - round, \(I=\frac{1}{2}MR^{2}\), that's correct. For the son, \(I = mr^{2}\), where \(r = 0.5\space m\), correct.

Wait, let's recalculate the angular acceleration: \(\tau = 55\times1.2=66\), \(I=\frac{1}{2}\times160\times(1.2)^{2}+35\times(0.5)^{2}=96\times1.44 + 8.75=138.24+8.75 = 146.99\). \(\alpha=\frac{66}{146.99}\approx0.449\space rad/s^{2}\). Total time \(t = 3\times1.4 = 4.2\space s\). \(\omega=\alpha t=0.449\times4.2\approx1.886\space rad/s\). Then \(v=\omega r\), \(r = 0.5\space m\), so \(v = 1.886\times0.5 = 0.943\space m/s\). But let's check if we should use the radius of the merry - go - round for the son's tangential speed? No, the son is at \(r = 0.5\space m\), so \(v=\omega r\) with \(r = 0.5\) is correct.

Wait, maybe we made a mistake in the moment of inertia. Let's recalculate \(I\):

\(I_{mg}=\frac{1}{2}\times160\times(1.2)^{2}=\frac{160\times1.44}{2}=115.2\space kg\cdot m^{2}\) (Oh! I made a mistake here earlier. \(\frac{1}{2}\times160 = 80\), \(80\times1.44 = 115.2\), not 138.24. That was the error!)

So \(I_{mg}=115.2\space kg\cdot m^{2}\), \(I_{son}=35\times(0.5)^{2}=8.75\space kg\cdot m^{2}\)

Total \(I=115.2 + 8.75=123.95\space kg\cdot m^{2}\approx124\space kg\cdot m^{2}\)

Now, recalculate \(\alpha=\frac{\tau}{I}=\frac{66}{123.95}\approx0.532\space rad/s^{2}\)

Then \(\omega=\alpha T=0.532\times4.2\approx2.234\space rad/s\)

Then \(v=\omega r=2.234\times0.5 = 1.117\space m/s\approx1.12\space m/s\)

Wait, let's do the moment of inertia of the merry - go - round again: \(M = 160\space kg\), \(R = 1.2\space m\), \(I=\frac{1}{2}MR^{2}=\frac{1}{2}\times160\times1.44 = 80\times1.44 = 115.2\space kg\cdot m^{2}\). That's correct. I had a mistake earlier in the first calculation of \(I_{mg}\) (multiplied 96 instead of 80).

So let's redo the steps:

1. Moment of inertia:
\(I_{mg}=\frac{1}{2}\times160\times(1.2)^{2}=115.2\space kg\cdot m^{2}\)
\(I_{son}=35\times(0.5)^{2}=8.75\space kg\cdot m^{2}\)
\(I = 115.2+8.75 = 123.95\space kg\cdot m^{2}\)

2. Torque: \(\tau=F\times R=55\times1.2 = 66\space N\cdot m\)

3. Angular acceleration: \(\alpha=\frac{\tau}{I}=\frac{66}{123.95}\approx0.532\space rad/s^{2}\)

4. Total time: \(t = 3\times1.4=4.2\space s\)

5. Angular velocity: \(\omega=\alpha t=0.532\times4.2\approx2.234\space rad/s\)

6. Tangential speed of son: \(v=\omega\times r\), where \(r = 0.5\space m\)
\(v = 2.234\times0.5=1.117\space m/s\approx1.12\space m/s\)

Wait, another way: Let's consider that the force is applied tangentially, so the torque is \(F\times R\), where \(R\) is the radius of the merry - go - round. The moment of inertia of the system is \(I=\frac{1}{2}MR^{2}+mr^{2}\). Let's use more precise calculations.

\(M = 160\), \(R = 1.2\), \(m = 35\), \(r = 0.5\), \(F = 55\), \(t_{total}=3\times1.4 = 4.2\)

\(I=\frac{1}{2}\times160\times1.2^{2}+35\times0.5^{2}=80\times1.44 + 35\times0.25=115.2+8.75 = 123.95\)

\(\tau = F\times R=55\times1.2 = 66\)

\(\alpha=\frac{\tau}{I}=\frac{66}{123.95}\approx0.5324\space rad/s^{2}\)

\(\omega=\alpha\times t_{total}=0.5324\times4.2\approx2.236\space rad/s\)

\(v=\omega\times r=2.236\times0.5 = 1.118\space m/s\approx1.12\space m/s\)

Alternatively, maybe we should use the radius of the merry - go - round for the torque, which is correct, and the son's distance from the center for his moment of inertia.

Another check: Let's calculate the angular acceleration again. \(\tau = 66\), \(I = 123.95\), \(\alpha=\frac{66}{123.95}\approx0.532\). Time is \(4.2\) seconds. \(\omega = 0.532\times4.2 = 2.234\). Then \(v=\omega\times0.5 = 1.117\space m/s\), approximately \(1.12\space m/s\).

Answer:

\(\approx1.12\space m/s\) (or more precisely, after re - calculating, if we use more decimal places:

\(\alpha=\frac{55\times1.2}{\frac{1}{2}\times160\times1.2^{2}+35\times0.5^{2}}=\frac{66}{115.2 + 8.75}=\frac{66}{123.95}\approx0.5324\)

\(\omega=0.5324\times3\times1.4 = 0.5324\times4.2\approx2.236\)

\(v = 2.236\times0.5=1.118\space m/s\approx1.12\space m/s\))