Question

A 50.0-kg child stands at the rim of a merry-go-round of radius 2.90 m, rotating with an angular speed of 3.70 rad/s.

(a) What is the child's centripetal acceleration?
m/s2
(b) What is the minimum force between her feet and the floor of the carousel that is required to keep her in the circular path?
N
(c) What minimum coefficient of static friction is required?
Is the answer you found reasonable? In other words, is she likely to stay on the merry-go-round? Yes or No

Answer 1

(a) The child's centripetal acceleration is 48.1 m/s².

(b) The minimum force between her feet and the floor of the carousel required to keep her in the circular path is 705 N.

(c) The minimum coefficient of static friction required is approximately 0.717.

Yes, she is likely to stay on the merry-go-round given that the calculated values meet the required forces for circular motion.

Step-by-step explanation:

(a) To find the centripetal acceleration of the child, we use the formula for centripetal acceleration:
`\(a_c = \frac{{v^2}}{r}\)`, where v is the tangential velocity and r is the radius. Given the angular speed
`\(\omega = 3.70 \, \text{rad/s}\)` and radius
`\(r = 2.90 \, \text{m}\)`, the tangential velocity
`\(v = r \cdot \omega = 2.90 \, \text{m} \cdot \, 3.70 \, \text{rad/s} = 10.73 \, \text{m/s}\)`. Substituting these values into the formula gives us
`\(a_c = \frac{{(10.73 \, \text{m/s})^2}}{2.90 \, \text{m}} = 48.1 \, \text{m/s}^2\)`.

(b) The minimum force keeping the child in the circular path is the centripetal force, which is provided by the normal force between her feet and the floor of the carousel. Using the formula
`\(F_c = m \cdot a_c\)`, where m is the mass and
`\(a_c\)` is the centripetal acceleration, we find
`\(F_c = 50.0 \, \text{kg} \cdot 48.1 \, \text{m/s}^2 = 705 \, \text{N}\)`.

(c) To determine the minimum coefficient of static friction needed, we compare the required centripetal force with the maximum frictional force:
`\(F_{\text{friction}} = \mu_{\text{static}} \cdot N\)`, where N is the normal force. The normal force equals the centripetal force in this scenario, so
`\(\mu_{\text{static}} = \frac{{F_c}}{{N}} = \frac{{705 \, \text{N}}}{{705 \, \text{N}}} = 0.717\)`. Therefore, the minimum coefficient of static friction required is approximately 0.717, ensuring she stays on the merry-go-round.

Answer 2

The centripetal acceleration of the child is approximately 39.826 m/s². The minimum force required to keep the child in the circular path is approximately 1991.3 N. The minimum coefficient of static friction required is approximately 4.08, which is not reasonable.

Step-by-step explanation:

(a) Centripetal acceleration:

To calculate the centripetal acceleration of the child, we can use the formula:
ac = r * ω²
where ac is the centripetal acceleration, r is the radius, and ω is the angular speed.
Substituting the given values:
ac = 2.90 m * (3.70 rad/s)²
ac ≈ 39.826 m/s²

(b) Minimum force:

The minimum force required to keep the child in the circular path is the centripetal force. The centripetal force can be calculated using the formula:
Fc = m * ac
where Fc is the centripetal force, m is the mass, and ac is the centripetal acceleration.
Substituting the given values:
Fc = (50.0 kg) * (39.826 m/s²)
Fc ≈ 1991.3 N

(c) Minimum coefficient of static friction:

The minimum coefficient of static friction required can be found using the formula:
µs = Ff / Fn
where µs is the coefficient of static friction, Ff is the frictional force, and Fn is the normal force.
Since the frictional force is the same as the centripetal force, we can use the value of Fc calculated in part (b).
Substituting the given values:
µs = 1991.3 N / (50.0 kg * 9.8 m/s²)
µs ≈ 4.08

The answer found is not reasonable as the coefficient of static friction cannot be greater than 1. Therefore, the child is likely to not stay on the merry-go-round.