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Chapter 10: Problem 51

A child with mass 40 kg sits on the edge of a merrygo-round at a distance of \(3.0 \mathrm{m}\) from its axis of rotation. The merry-go-round accelerates from rest up to \(0.4 \mathrm{rev} / \mathrm{s}\) in 10 s. If the coefficient of static friction between the child and the surface of the merry-go-round is 0.6, does the child fall off before \(5 \mathrm{s} ?\)

Short Answer

Expert verified
The child does not fall off before 5 seconds, as the centripetal force they experience at 5 seconds (50.268 N) is less than the maximum centripetal force they can withstand without falling (235.2 N).

Step by step solution

01

Identify the given quantities

We are given the following quantities: - Mass of the child (m): 40 kg - Distance from the axis of rotation (r): 3.0 m - Final rotational velocity (\(v_f\)): 0.4 rev/s - Time to reach the final velocity (t): 10 s - Coefficient of static friction (\(\mu_s\)): 0.6
02

Calculate the maximum centripetal force without falling

To determine the maximum centripetal force without falling, we'll need to find the maximum static friction force. Using the formula for static friction, we have: \(F_s = \mu_s F_N\) where \(F_s\) is the static friction force and \(F_N\) is the normal force. Since the child is in contact with a horizontal surface, the normal force equals the gravitational force acting on the child: \(F_N = mg\) Substituting the values, we get: \(F_s = (0.6)(40\ kg)(9.8\ \mathrm{m/s^2})\) \(F_s = 235.2\ N\) The maximum centripetal force without falling is 235.2 N.
03

Calculate the centripetal acceleration at 5 seconds

First, we need to find the linear acceleration (\(a\)) of the child. We know the final velocity and time, so we can use the formula: \(v_f = a \times t\) We need to convert the final velocity from rev/s to m/s by multiplying it with the circumference of the circle: \(v_f = 0.4\ \text{rev/s} \times 2\pi (3 \text{m}) = 7.54\ \mathrm{m/s}\) Now, we can find the acceleration: \(a = \frac{v_f}{t} = \frac{7.54\ \mathrm{m/s}}{10\ \mathrm{s}}\) \(a = 0.754\ \mathrm{m/s^2}\) Now, we can find the centripetal acceleration at 5 seconds: \(a_c = \frac{a \times 5\ \mathrm{s}}{3\ \mathrm{m}}\) \(a_c = 1.2567\ \mathrm{m/s^2}\) The centripetal acceleration at 5 seconds is 1.2567 m/s虏.
04

Calculate the centripetal force at 5 seconds

Now that we have the centripetal acceleration, we can find the centripetal force using the formula: \(F_c = ma_c\) Substituting the given values: \(F_c = (40\ kg)(1.2567\ \mathrm{m/s^2})\) \(F_c = 50.268\ N\) The centripetal force at 5 seconds is 50.268 N.
05

Check if the child falls off

Now, we compare the centripetal force at 5 seconds to the maximum centripetal force without falling. If the force at 5 seconds is greater than the maximum, the child will fall off: \(50.268\ N > 235.2\ N\) This inequality is false; therefore, the child does not fall off before 5 seconds.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Static Friction and Safety on the Merry-Go-Round
Static friction is the force that helps keep objects stationary relative to each other. It acts between the surfaces at rest, like the child and the merry-go-round. In this scenario, the static friction force is crucial in preventing the child from sliding off. The formula for static friction is given by:
  • \(F_s = \mu_s F_N\)
where \(F_s\) is the static friction force, and \(\mu_s\) is the coefficient of static friction. The normal force \(F_N\) here is equivalent to the gravitational force \(mg\).
This force must be greater than or equal to the centripetal force needed to keep the child from falling. If the friction is surpassed, the child might slide off. Therefore, calculating the maximum static friction helps determine the safety of the situation.
Understanding Rotational Motion
Rotational motion involves objects moving in a circular path. The merry-go-round is a perfect example of rotational motion. The speed at which it rotates is given in revolutions per second, but we often need this in meters per second to solve physics problems.
To convert, we use the circle's circumference formula since one revolution equals one full circle:
  • \(v_f = 0.4 \, \text{rev/s} \times 2\pi (3 \, \text{m}) \)
This conversion helps in calculating linear velocity and, subsequently, the acceleration. Understanding these conversions and the nature of circular motion forms a fundamental part of analyzing any rotating system.
Centripetal Acceleration in Circular Paths
Centripetal acceleration is essential for maintaining an object's path in a circle. It points towards the center of the circle and keeps the object moving along its curved path. It's calculated using the formula:
  • \(a_c = \frac{v^2}{r} \)
However, to find the acceleration at a specific time, we often work with the equation \(a = \frac{v_f}{t}\) for linear acceleration and adjust it for centripetal cases.
In the problem, knowing the centripetal acceleration is vital to calculate if the static friction is enough to prevent the child from slipping. Understanding these physics principles allows for better analysis of forces at play in any rotational scenario.

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Most popular questions from this chapter

Zorch, an archenemy of Rotation Man, decides to slow Earth's rotation to once per \(28.0 \mathrm{h}\) by exerting an opposing force at and parallel to the equator. Rotation Man is not immediately concerned, because he knows Zorch can only exert a force of \(4.00 \times 10^{7} \mathrm{N}\) (a little greater than a Saturn V rocket's thrust). How long must Zorch push with this force to accomplish his goal? (This period gives Rotation Man time to devote to other villains.)

A vertical wheel with a diameter of \(50 \mathrm{cm}\) starts from rest and rotates with a constant angular acceleration of \(5.0 \mathrm{rad} / \mathrm{s}^{2}\) around a fixed axis through its center counterclockwise. (a) Where is the point that is initially at the bottom of the wheel at \(t=10 \mathrm{s} ?\) (b) What is the point's linear acceleration at this instant?

An aircraft is coming in for a landing at 300 meters height when the propeller falls off. The aircraft is flying at \(40.0 \mathrm{m} / \mathrm{s}\) horizontally. The propeller has a rotation rate of \(20 \mathrm{rev} / \mathrm{s},\) a moment of inertia of \(70.0 \mathrm{kg}-\mathrm{m}^{2},\) and a mass of 200 kg. Neglect air resistance. (a) With what translational velocity does the propeller hit the ground? (b) What is the rotation rate of the propeller at impact?

When reducing the mass of a racing bike, the greatest benefit is realized from reducing the mass of the tires and wheel rims. Why does this allow a racer to achieve greater accelerations than would an identical reduction in the mass of the bicycle's frame?

An automobile engine can produce \(200 \mathrm{N} \cdot \mathrm{m}\) of torque. Calculate the angular acceleration produced if \(95.0 \%\) of this torque is applied to the drive shaft, axle, and rear wheels of a car, given the following information. The car is suspended so that the wheels can turn freely. Each wheel acts like a 15.0-kg disk that has a 0.180-m radius. The walls of each tire act like a 2.00-kg annular ring that has inside radius of \(0.180 \mathrm{m}\) and outside radius of 0.320 m. The tread of each tire acts like a 10.0 -kg hoop of radius \(0.330 \mathrm{m} .\) The 14.0-kg axle acts like a rod that has a \(2.00-\mathrm{cm}\) radius. The 30.0-kg drive shaft acts like a rod that has a 3.20-cm radius.
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