Question

Consider a father pushing a child on a playground merry-go-round. The system has a moment of inertia of 84.4 kg · m2. The father exerts a force on the merry-go-round perpendicular to its 1.50 m radius to achieve a torque of 375 N · m.(a) Calculate the rotational kinetic energy (in J) in the merry-go-round plus child when they have an angular velocity of 29.2 rpm. J(b) Using energy considerations, find the number of revolutions the father will have to push to achieve this angular velocity starting from rest. revolutions(c) Again, using energy considerations, calculate the force (in N) the father must exert to stop the merry-go-round in seven revolutions. N

Answer 1

Answer:

A)The rotational kinetic energy is ≈ 9.9962 J

B)The number of revolutions to achieve that angular velocity = 0.0085

C)Force to stop it = 0.15152N

Explanation:

Given :

• The system has a moment of inertia of $84.4kgm^{2}$
• Radius $R = 1.50m$
• Torque exerted = $375Nm$

A)

Rotational kinetic energy :

$E = \frac{1}{2} Iw^{2}$

where,

• $E$ is the rotational kinetic energy $(J)$
• $I$ is the moment of inertia $(Kgm^{2})$
• $w$ is the angular velocity $(rads^{-1})$

⇒$I=84.4\\w=29.2rpm=\frac{29.2}{60} rps=0.4867 rads^{-1}$

⇒$E = \frac{1}{2} Iw^{2}\\E=\frac{1}{2} *84.4*(0.4867)^{2}\\E=9.9962 J$

Thus the energy is ≈ 9.9962 J

B) Work done by the torque = The change in rotational kinetic energy

 $T\alpha = \frac{1}{2} Iw^{2} - \frac{1}{2} Iw_{o}^{2}$

where ,

• $T$ is the torque exerted
• $\alpha$ is the angle traversed
• $w_o$ is the initial angular velocity.

Thus,

$375*\alpha =\frac{1}{2} *84.4*(0.4867)^{2}-\frac{1}{2} *84.4*0\\\alpha =0.026657 rad\\=\frac{0.026657}{\pi } rev\\=0.0085 rev$

The number of revolutions = 0.0085

C)

7 revolutions :

$=7*2\pirad\\=14\pirad$

 $T\alpha = \frac{1}{2} Iw^{2} - \frac{1}{2} Iw_{o}^{2}$

$T*14\pi  =9.9962\\T= 0.2273 Nm\\F=\frac{T}{R} =\frac{0.2273}{1.5} \\F=0.15152N$

Force = 0.15152N