Calculating Force and Acceleration for a Yoyo and Clock

In summary: That seems reasonable. In summary, a weight on the end of a string spinning around a fixed axle at a velocity of 30m/s has a force of .00328 N.
  • #1
Julia Darko
9
0

Homework Statement


A yoyo is spun with a radius of 0.5m. It reaches a velocity of 30m/s. What is the force if the yoyo is 0.3kg?
----
A clock has a hand with 0.3m long and 0.5kg. It takes 60 seconds for 1 revolution. What is its acceleration? What is the force?

If you have any tips or instructions, please comment,

Homework Equations


Ac= v^2/r
2(3.14)r/t
F=ma

The Attempt at a Solution


I attempted to solve the first one and I got 1800 m/s and 540 Newtons, but I feel this is wayyyyyyyyyyyyyy off. Please save my soul someone.
 
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  • #2
Julia Darko said:

Homework Statement


A yoyo is spun with a radius of 0.5m. It reaches a velocity of 30m/s. What is the force if the yoyo is 0.3kg?
----
A clock has a hand with 0.3m long and 0.5kg. It takes 60 seconds for 1 revolution. What is its acceleration? What is the force?

If you have any tips or instructions, please comment,

Homework Equations


Ac= v^2/r
2(3.14)r/t
F=ma

The Attempt at a Solution


I attempted to solve the first one and I got 1800 m/s and 540 Newtons, but I feel this is wayyyyyyyyyyyyyy off. Please save my soul someone.
The 540 N looks to be in the right ballpark, so that's not so worrying. To say more we'll have to see the details of how you arrived at that value.

I'm not sure what the 1800 m/s could pertain to. It doesn't seem to be an answer to a question posed in your problem statement.
 
  • #3
For my first equation, I plugged in A=(30)^2/.5 and got 1800 m/s^2. Then I used F=ma... F=(.3)(1800) and got 540N. I wasn't there when this was taught so I'm not sure if I'm heading in the right direction. :/
 
  • #4
Julia Darko said:
A yoyo is spun with a radius of 0.5m. It reaches a velocity of 30m/s. What is the force if the yoyo is 0.3kg?
This is not at all clear. A yo-yo has two radii - its outer radius and the radius of its axle.
What force? Is this perhaps the tension in the string as the yo-yo descends under gravity, when it has reached 30m/s?

Edit: Oh I see, it may be a yo-yo but it's being treated as just mass on the end of a string of fixed length. (Are you sure that's the right interpretation? It's not what is usually meant by a spinning yo-yo.)
But is this in the presence of gravity? If so, for the force to be constant it will be spun in a horizontal plane, and gravity will affect the tension. So my question becomes whether the force is supposed to be just the centripetal force or the whole tension.
 
Last edited:
  • #5
Julia Darko said:
For my first equation, I plugged in A=(30)^2/.5 and got 1800 m/s^2. Then I used F=ma... F=(.3)(1800) and got 540N. I wasn't there when this was taught so I'm not sure if I'm heading in the right direction. :/
Okay, that's fine and makes sense now that the details are there. Your results are fine.

Now, what's your approach for the second question?
 
  • #6
haruspex said:
This is not at all clear. A yo-yo has two radii - its outer radius and the radius of its axle.
What force? Is this perhaps the tension in the string as the yo-yo descends under gravity, when it has reached 30m/s?
I believe that the yo-yo in this case just represents a weight on the end of a string, said weight undergoing circular motion with the string as the radius.
 
  • #7
gneill said:
I believe that the yo-yo in this case just represents a weight on the end of a string, said weight undergoing circular motion with the string as the radius.
Yes, I realized just after your previous post that that's how it was being interpreted. But please see my follow-up question in post #4.
 
  • #8
gneill said:
Okay, that's fine and makes sense now that the details are there. Your results are fine.

Now, what's your approach for the second question?
Well for the second equation I just used the 2πR/T, so 2(3.14)(.3)/60seconds=.0314m/s. Then I plugged that into A=v^2/r, A=.0314^2/.3 and got .00328 m/s^2. Then, I plugged that it F=ma.. So F=(.5)(.00328)= .00164 N.
 
  • #9
Julia Darko said:
Well for the second equation I just used the 2πR/T, so 2(3.14)(.3)/60seconds=.0314m/s. Then I plugged that into A=v^2/r, A=.0314^2/.3 and got .00328 m/s^2. Then, I plugged that it F=ma.. So F=(.5)(.00328)= .00164 N.
Okay, the calculations look fine. You may want to make sure that the number of significant figures in your final results correspond to the number of significant figures in the given data.
 

1. How do I calculate the force and acceleration of a yoyo or clock?

To calculate the force and acceleration of a yoyo or clock, you will need to use Newton's second law of motion. This law states that the force applied to an object is equal to the mass of the object multiplied by its acceleration. In this case, the force will be the tension in the string of the yoyo or the force exerted by the clock's mechanism, and the acceleration will be the change in velocity over time.

2. What is the formula for calculating force and acceleration?

The formula for calculating force and acceleration is F = ma, where F is the force, m is the mass of the object, and a is the acceleration. This formula can be used for both the yoyo and the clock, as long as you have the necessary values for mass and acceleration.

3. How do I measure the mass of a yoyo or clock?

To measure the mass of a yoyo or clock, you can use a scale. Place the object on the scale and record the measurement in kilograms. If you do not have a scale, you can estimate the mass based on the size and material of the object.

4. What units should I use for force and acceleration?

The standard units for force are Newtons (N) and for acceleration are meters per second squared (m/s^2). However, you may also see force measured in kilograms times meters per second squared (kg*m/s^2), which is equivalent to a Newton.

5. Can I use the same formula for calculating force and acceleration for a yoyo and a clock?

Yes, you can use the same formula, F = ma, for calculating force and acceleration for both a yoyo and a clock. However, the mass and acceleration values will be different for each object, so make sure to use the correct values in the formula for accurate calculations.

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