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A machine part has the shape of a solid uniform sphere of mass 225 g and diameter 3.00 cm. It is spinning about a frictionless axle through its center, but at one point on its equator it is scraping against metal, resulting in a friction force of 0.0200 N at that point. (a) Find its angular acceleration. (b) How long will it take to decrease its rotational speed by 22.5 rad/s ?

Short Answer

Expert verified
Angular acceleration is 66.67 rad/s²; time to decrease speed is 0.337 s.

Step by step solution

01

Determine the Moment of Inertia

To find the angular acceleration, we first need to find the moment of inertia of the sphere. The formula for the moment of inertia \( I \) of a solid sphere about an axle through its center is \( I = \frac{2}{5}mr^2 \), where \( m \) is the mass and \( r \) is the radius. The mass \( m \) is 225 g or 0.225 kg and the diameter is 3.00 cm, so the radius \( r \) is 1.50 cm or 0.015 m.Plug these numbers into the formula:\[ I = \frac{2}{5}(0.225)(0.015)^2 \]Calculate the value:\[ I = \frac{2}{5}(0.225)(0.000225) = 4.5 \times 10^{-6} \text{ kg} \cdot \text{m}^2 \]
02

Find Angular Acceleration

Use Newton’s second law for rotation, \( \tau = I\alpha \), where \( \tau \) is the torque and \( \alpha \) is the angular acceleration. The torque \( \tau \) can be calculated by the friction force \( f \) and the radius \( r \):\( \tau = fr = 0.0200 \times 0.015 \).Calculate the torque:\[ \tau = 3 \times 10^{-4} \text{ N} \cdot \text{m} \]Then, find the angular acceleration \( \alpha \):\[ \alpha = \frac{\tau}{I} = \frac{3 \times 10^{-4}}{4.5 \times 10^{-6}} = 66.67 \text{ rad/s}^2 \]
03

Determine Time to Decrease Rotational Speed

Using the equation for angular motion, \( \omega_f = \omega_i + \alpha t \), where \( \omega_f \) is the final angular velocity, \( \omega_i \) is the initial angular velocity, \( \alpha \) is the angular acceleration, and \( t \) is the time. We know \( \omega_f = \omega_i - 22.5 \text{ rad/s} \).Rearrange this equation to solve for time:\[ t = \frac{\Delta \omega}{\alpha} = \frac{22.5}{66.67} = 0.337 \text{ s} \]
04

Final Check and Summary

Verify the calculations and ensure the units are consistent throughout the problem. The calculations give an angular acceleration \( \alpha = 66.67 \text{ rad/s}^2 \) and a time to decrease the speed by 22.5 rad/s of \( t = 0.337 \text{ s} \).

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Key Concepts

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

Moment of Inertia
The concept of moment of inertia is crucial in understanding how objects spin. Think of it as the rotational equivalent of mass in linear motion. It's a measure of how difficult it is to change the rotational speed of an object. For different shapes, the formulas for calculating moment of inertia vary.

In the exercise, we determine the moment of inertia for a solid sphere using the formula:
  • \( I = \frac{2}{5}mr^2 \)
Here, \( m \) is the mass of the sphere, and \( r \) is the radius. This formula reflects the fact that mass distributed further from the axis makes an object harder to start or stop spinning.

For our problem, we've converted the mass to kilograms and radius to meters to use consistent units. By substituting these values into the formula, we find the sphere’s moment of inertia is \( 4.5 \times 10^{-6} \, \text{kg} \cdot \text{m}^2 \). Understanding and calculating moment of inertia helps in predicting how a given force will affect an object’s rotational motion.
Rotational Motion Equations
Rotational motion equations describe the behavior of rotating objects. They're similar to equations for linear motion but applied to angles and rotations.

One important equation used in this scenario is Newton's second law for rotation:
  • \( \tau = I\alpha \)
Here, \( \tau \) is the torque, \( I \) is the moment of inertia, and \( \alpha \) is the angular acceleration. This formula shows how torque affects rotational acceleration, much like force affects linear acceleration.

Another key equation used is for calculating changes in angular speed over time:
  • \( \omega_f = \omega_i + \alpha t \)
This helps determine how the angular velocity \( \omega \) changes with time \( t \). In this exercise, by rearranging the equation, we calculated the time it takes for the sphere’s rotational speed to decrease by a specific amount, given its angular acceleration.

These equations are critical tools in solving problems involving rotational dynamics.
Torque
Torque is the rotational force that causes an object to spin. It plays a role analogous to linear force in motion. Essentially, torque determines how much a force acting at a distance will rotate an object.

It is calculated as:
  • \( \tau = fr \)
where \( f \) is the force acting perpendicular to the point of rotation, and \( r \) is the distance from the axis of rotation. This exercise involved calculating torque using the frictional force and the radius of the sphere.

The calculation\( \tau = 3 \times 10^{-4} \text{ N} \cdot \text{m} \) gives us the torque resulting from friction. With this torque, we showed how it affects the sphere's angular acceleration using Newton’s second law for rotation.

Torque is fundamental in understanding and controlling rotational movements in mechanical systems.

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Most popular questions from this chapter

One force acting on a machine part is \(\overrightarrow{F} = (-5.00 N)\hat{\imath} + (4.00 N)\hat{\jmath}\). The vector from the origin to the point where the force is applied is \(\overrightarrow{r} = (-0.450 m)\hat{\imath} + (0.150 m)\hat{\jmath}\). (a) In a sketch,show \(\overrightarrow{r}, \overrightarrow{F},\) and the origin. (b) Use the right- hand rule to determine the direction of the torque. (c) Calculate the vector torque for an axis at the origin produced by this force. Verify that the direction of the torque is the same as you obtained in part (b).

A size-5 soccer ball of diameter 22.6 cm and mass 426 g rolls up a hill without slipping, reaching a maximum height of 5.00 m above the base of the hill. We can model this ball as a thin-walled hollow sphere. (a) At what rate was it rotating at the base of the hill? (b) How much rotational kinetic energy did it have then?

The solid wood door of a gymnasium is 1.00 m wide and 2.00 m high, has total mass 35.0 kg, and is hinged along one side. The door is open and at rest when a stray basketball hits the center of the door head-on, applying an average force of 1500 N to the door for 8.00 ms. Find the angular speed of the door after the impact. [\(Hint:\) Integrating Eq. (10.29) yields \(\Delta L_z = \int_{t1}^{t2} (\Sigma\tau_z)dt = (\Sigma\tau_z)_av \Delta t\). The quantity \(\int_{t1}^{t2} (\Sigma\tau_z)dt\) is called the angular impulse.]

A thin, uniform, \(3.80-\mathrm{~kg}\) bar, \(80.0 \mathrm{~cm}\) long, has very small \(2.50-\mathrm{~kg}\) balls glued on at either end (Fig. \(\mathbf{P 1 0 . 5 7}\) ). It is supported horizontally by a thin, horizontal, frictionless axle passing through its center and perpendicular to the bar. Suddenly the right- hand ball becomes detached and falls off, but the other ball remains glued to the bar. (a) Find the angular acceleration of the bar just after the ball falls off. (b) Will the angular acceleration remain constant as the bar continues to swing? If not, will it increase or decrease? (c) Find the angular velocity of the bar just as it swings through its vertical position.

A uniform marble rolls down a symmetrical bowl, starting from rest at the top of the left side. The top of each side is a distance \(h\) above the bottom of the bowl. The left half of the bowl is rough enough to cause the marble to roll without slipping, but the right half has no friction because it is coated with oil. (a) How far up the smooth side will the marble go, measured vertically from the bottom? (b) How high would the marble go if both sides were as rough as the left side? (c) How do you account for the fact that the marble goes \(higher\) with friction on the right side than without friction?

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