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A machine part has the shape of a solid uniform sphere of mass 225 \(\mathrm{g}\) and diameter 3.00 \(\mathrm{cm} .\) 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 \(\mathrm{N}\) at that point. (a) Find its angular acceleration. (b) How long will it take to decrease its rotational speed by 22.5 \(\mathrm{rad} / \mathrm{s} ?\)

Short Answer

Expert verified
(a) Angular acceleration is 22.22 rad/s². (b) Time is 1.01 seconds.

Step by step solution

01

Convert mass and diameter to standard units

The mass of the sphere is given as 225 g, which needs to be converted to kilograms. Since 1 kg = 1000 g, we have:\[ \text{Mass} (m) = \frac{225}{1000} \text{ kg} = 0.225 \text{ kg}. \]The diameter is given as 3.00 cm. To find the radius (r) in meters, divide by 2 and convert centimeters to meters:\[ r = \frac{3.00}{2 \times 100} \text{ m} = 0.015 \text{ m}. \]
02

Find the moment of inertia of the sphere

The formula for the moment of inertia (I) of a solid sphere about its center is:\[ I = \frac{2}{5} m r^2. \]Substitute the mass and radius found in Step 1 into this formula:\[ I = \frac{2}{5} \times 0.225 \text{ kg} \times (0.015 \text{ m})^2 = 1.35 \times 10^{-5} \text{ kg} \cdot \text{m}^2. \]
03

Calculate the torque due to friction

Torque (Ï„) is given by the formula:\[ \tau = F \times r, \]where F is the friction force and r is the radius. Here, F = 0.0200 N and r = 0.015 m:\[ \tau = 0.0200 \text{ N} \times 0.015 \text{ m} = 3.00 \times 10^{-4} \text{ N} \cdot \text{m}. \]
04

Determine the angular acceleration

Angular acceleration (α) is given by the formula:\[ \alpha = \frac{\tau}{I}, \]where τ is the torque and I is the moment of inertia. Substitute the values from Steps 2 and 3:\[ \alpha = \frac{3.00 \times 10^{-4} \text{ N} \cdot \text{m}}{1.35 \times 10^{-5} \text{ kg} \cdot \text{m}^2} \approx 22.22 \text{ rad/s}^2. \]
05

Calculate the time to change rotational speed

Use the kinematic equation for angular motion:\[ \Delta \omega = \alpha \times t, \]where \( \Delta \omega = 22.5 \text{ rad/s} \) and \( \alpha \approx 22.22 \text{ rad/s}^2. \)Solve for time (t):\[ t = \frac{\Delta \omega}{\alpha} = \frac{22.5 \text{ rad/s}}{22.22 \text{ rad/s}^2} \approx 1.01 \text{ s}. \]

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Key Concepts

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

Angular Acceleration
Angular acceleration (\( \alpha \)) describes how quickly an object's rotational speed changes. It's essentially the rotational equivalent of linear acceleration. In the exercise you're looking at, the machine part is subject to friction as it spins, which affects its speed.

In our context, angular acceleration tells us how fast the speed of the sphere spinning around its axle is decreasing due to the friction. To calculate this, we apply the formula:
  • \( \alpha = \frac{\tau}{I} \)
Here, \( \tau \) represents the torque applied to the object and \( I \) is the object's moment of inertia.

The negative value of angular acceleration here would indicate a deceleration, because friction is slowing down the rotation. Understanding this decrease is crucial for predicting how long it takes for the sphere's rotation to diminish to a certain speed.
Moment of Inertia
The moment of inertia (\( I \)) is a property of a body that defines its resistance to angular acceleration. Think of it as the rotational equivalent of mass in linear dynamics. It tells you how much torque is needed to achieve a certain angular acceleration.

For this solid sphere, the formula used is:
  • \( I = \frac{2}{5}m r^2 \)
This depends on both the mass of the sphere (\( m \)) and its radius (\( r \)). Knowing the moment of inertia helps calculate how the sphere will react to the frictional force trying to slow it down. A larger moment means more resistance.

This concept helps us determine the amount of torque necessary to change the spinning speed of the sphere, considering its mass distribution relative to its axis of rotation.
Torque
Torque (\( \tau \)) is the force that causes an object to rotate. For linear motion, we have force, and similarly, for rotational motion, we have torque. It essentially causes a turn or twist about an axis.

In the exercise, torque is produced by the frictional force acting at the equator of the spinning sphere. It's calculated using the formula:
  • \( \tau = F \times r \)
where \( F \) is the force of friction and \( r \) is the radius of the sphere.

In simpler terms, the further the force is applied from the axis, the larger its effect (or leverage) and the greater the torque. Understanding torque is vital in finding how this acting friction impacts the sphere's rotation speed, leading to angular deceleration.
Friction Force
Friction force is the force that resists the motion of one surface against another. It's the 'drag' that prevents surfaces from sliding past each other without effort. Even when it is a small force, as in our exercise, it has a significant role.

Here, the sphere's edge rubs against metal, creating a friction force of 0.0200 N. This friction is the cause behind the torque that decelerates the sphere, thereby inducing angular acceleration in the opposite direction of its initial spin.
  • Friction opposes motion; it converts kinetic energy into heat.
  • It's responsible for slowing down moving parts, which is crucial in real-world applications.
By grasping the basics of friction force, you can better understand how it affects rotational dynamics beyond the surface level influence on wheels and spheres.

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Most popular questions from this chapter

The Yo-yo. A yo-yo is made from two uniform disks, each with mass \(m\) and radius \(R\) , connected by a light axle of radius \(b .\) A light, thin string is wound several times around the axle and then held stationary while the yo-yo is released from rest, dropping as the string unwinds. Find the linear acceleration and angular acceleration of the yo-yo and the tension in the string.

A \(2.20-\) kg hoop 1.20 \(\mathrm{m}\) in diameter is rolling to the right without slipping on a horizontal floor at a steady 3.00 \(\mathrm{rad} / \mathrm{s}\) . (a) How fast is its center moving? (b) What is the total kinetic energy of the hoop? (c) Find the velocity vector of each of the following points, as viewed by a person at rest on the ground: (i) the highest point on the hoop; (ii) the lowest point on the hoop; (iii) a point on the right side of the hoop, midway between the top and the bottom. (d) Find the velocity vector for each of the points in part (c), but this time as viewed by someone moving along with the same velocity as the hoop.

One force acting on a machine part is \(\vec{\boldsymbol{F}}=(-5.00 \mathrm{N}) \hat{\imath}+\) \((4.00 \mathrm{N}) \hat{\boldsymbol{J}}\) . The vector from the origin to the point where the force is applied is \(\vec{r}=(-0.450 \mathrm{m}) \hat{\imath}+(0.150 \mathrm{m}) \hat{\boldsymbol{J}}\) (a) In a sketch, show \(\vec{r}, \vec{\boldsymbol{F}},\) 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 playground merry-go-round has radius 2.40 \(\mathrm{m}\) and moment of inertia 2100 \(\mathrm{kg} \cdot \mathrm{m}^{2}\) 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?

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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