Question

A playground merry-go-round has radius 2.40 m and moment of inertia 2100 kg⋅m2 about a vertical axle through its center, and it turns with negligible friction.
(a) A child applies an 18.0 N force tangentially to the edge of the merry-go-round for 15.0 s. If the merry-go-round is initially at rest, what is its angular speed after this 15.0 s interval?
(b) How much work did the child do on the merry-go-round?
(c) What is the average power supplied by the child?

Answer 1

Answer:

a) 0.31 rad/s

b) 100 J

c) 6.67 W

Explanation:

(a) the force would generate a torque of:

$T = FR = 18 * 2.4 = 43.2 Nm$

According to Newton 2nd law, the angular acceleration would be

$\alpha = \frac{T}{I} = \frac{43.2}{2100} = 0.021 rad/s^2$

It starts from rest, then after 15s it would achieve a speed of

$\omega = \alpha t = 0.021 * 15 = 0.31 rad/s$

(b) The distance angle swept by it is:

$\theta = \frac{\alpha t^2}{2} = \frac{0.021 * 15^2}{2} = 2.314 rad$

Hence the work by the child

$W = T\theta = 43.2 *2.314  \approx 100 J$

c) Average power to work per time unit

$P = \frac{W}{t} = \frac{100}{15} = 6.67 W$

Answer 2

The angular speed after a 15s interval would be 0.31 rad/s

The work done by the child on the merry-go-round is 100 J

The average power supplied by the child is 6.67 W

Calculations and Parameters:

The force would generate a torque of:

T= FR = 18 x 2.4= 43.2Nm

From the 2nd law of Newton, the angular acceleration would be

43.2/2100= 0.021 rad/s^2

Hence, if it starts from rest, then after 15s, the angular speed would be 0.021 x 15= 0.31 rad/s^2.

The distance would be 0.021 x 15^2/2

=>2. 314 rad.

The work done by the child would be 43.2 x 2.314 which is aprox 100J.

Average work per unit

P= W/t

= 100/5

= 6.67W

Read more about angular speed here:

https://brainly.com/question/6860269