# Speed and accleration of a turntable

• clope023
The formula for circumference is C=pid where d is the diameter. So 0.740m is the diameter, so to get the radius you divide it by 2 giving you 0.370m. So after 1 revolution, the blade would have traveled a distance of pi(0.370) = 1.164m. To find the tangential speed, you need to divide the distance by the time, which is 0.194s. So the tangential speed would be 1.164m/0.194s = 6.0m/s. For part D, you need to use the formula a = arad + atan, where arad is
clope023

## Homework Statement

An electric turntable 0.740m in diameter is rotating about a fixed axis with an initial angular velocity of 0.290rad/s . The angular acceleration is 0.900rad/s^2 .

a) Compute the angular velocity after a time of 0.194s. - .465rev/s

b) Through how many revolutions has the blade turned in this time interval? - .073rev

c) What is the tangential speed of a point on the tip of the blade at time t = 0.194s ?

d) What is the magnitude of the resultant acceleration of a point on the tip of the blade at time = 0.194 ?

## Homework Equations

v = r$$\omega$$

arad = $$\omega$$^2r

atan = r$$\alpha$$

## The Attempt at a Solution

for part c)

v = (.740/2)(.465) = .172m/s wrong

for part d)

atan = (.740/2)(.900) = .333m/s^2

a = (.0800^2 + .333^2)^(1/2) = .342m/s^2 wrong

I'm not sure what I'm doing wrong, any help is appreciated.

For part C it seems as if you need to do is convert the revs to meters. To do this you need to know the circumference (which would be the distance "unrolled" after 1 revolution)

Hello! I can help you with your questions about the speed and acceleration of a turntable. It seems like you have attempted to use the correct equations, but there may be some errors in your calculations. Let's go through each part together to find the correct answers.

a) To compute the angular velocity after a time of 0.194s, we can use the equation:

\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. Plugging in the given values, we get:

\omega_f = 0.290 + (0.900)(0.194) = 0.443 rad/s

b) To find the number of revolutions, we can use the equation:

\theta = \omega_it + \frac{1}{2}\alpha t^2

Where \theta is the angle of rotation, \omega_i is the initial angular velocity, \alpha is the angular acceleration, and t is the time. Plugging in the given values, we get:

\theta = (0.290)(0.194) + \frac{1}{2}(0.900)(0.194)^2 = 0.073 rev

c) To find the tangential speed at time t = 0.194s, we can use the equation:

v = r\omega

Where v is the tangential speed, r is the radius, and \omega is the angular velocity. Plugging in the given values, we get:

v = (0.740/2)(0.443) = 0.163 m/s

d) To find the magnitude of the resultant acceleration, we can use the Pythagorean theorem:

Where a is the magnitude of the resultant acceleration, a_{rad} is the radial acceleration, and a_{tan} is the tangential acceleration. Plugging in the given values, we get:

a = \sqrt{(0.900)^2(0.740/2)^2 + (0.740/2)(0.443)^2} = 0.393 m/s^2

I hope this helps clarify any confusion and provides you with the

## 1. What is the relationship between speed and acceleration on a turntable?

Speed and acceleration are directly related on a turntable. As the speed of the turntable increases, so does the acceleration. This means that the faster the turntable is spinning, the faster the objects on it will accelerate.

## 2. How does the diameter of the turntable affect its speed and acceleration?

The diameter of the turntable does not directly affect its speed and acceleration. However, a larger diameter turntable will have a higher linear speed at the outer edge compared to a smaller diameter turntable. This means that the acceleration at the outer edge will also be greater on a larger diameter turntable.

## 3. Can the speed and acceleration of a turntable be changed?

Yes, the speed and acceleration of a turntable can be changed. This can be done by adjusting the motor speed or by changing the friction between the turntable and the objects on it. By changing these factors, the speed and acceleration of the turntable can be increased or decreased.

## 4. What is the role of centripetal force in the speed and acceleration of a turntable?

Centripetal force is responsible for keeping objects on a turntable moving in a circular path. This force is also responsible for the change in direction of the objects, which creates acceleration. Therefore, centripetal force is a crucial factor in the speed and acceleration of a turntable.

## 5. How does the shape of the turntable affect its speed and acceleration?

The shape of the turntable does not have a significant impact on its speed and acceleration. As long as the turntable is circular and has a smooth surface, the speed and acceleration will be similar. However, a non-circular or unevenly shaped turntable may result in inconsistent speed and acceleration due to varying distances between the center and outer edge.

• Introductory Physics Homework Help
Replies
5
Views
2K
• Introductory Physics Homework Help
Replies
17
Views
2K
• Introductory Physics Homework Help
Replies
5
Views
5K
• Introductory Physics Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
5
Views
2K
• Introductory Physics Homework Help
Replies
9
Views
9K
• Introductory Physics Homework Help
Replies
5
Views
2K
• Introductory Physics Homework Help
Replies
16
Views
8K
• Introductory Physics Homework Help
Replies
6
Views
7K
• Introductory Physics Homework Help
Replies
2
Views
7K