Power and torque what did i do wrong?

In summary: The instantaneous power is found by taking the torque output at the instant of interest and multiplying it by the angular velocity. In this case, it is just (torque*radian)/2. So the instantaneous power would be (2690.8*2)/16s = 783.2Watts.
  • #1
ahello888a
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power and torque what did i do wrong?

Homework Statement


A rotating uniform cylindrical platform of mass 230 kg and radius 5.6 m slows down from 3.8 rev/s to rest in 16 s when the driving motor is disconnected. Estimate the power output of the motor (\rm hp) required to maintain a steady speed of 3.8 rev/s


Homework Equations


Wfriction=change in KE = -.5*I*w^2
power=work/time


The Attempt at a Solution


So I was was able to get the amount of work that the friction force does and it was -.5 (.5 * MR^2)w^2 = 45053.39J and that work done by friction is the same amount of work needed for the motor to output. So the power of the motor needed is 45053.39/16s = 2690.8Watts and to convert that to HP I just divide by 746W and got 3.6HP. is that right?
 
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  • #2


I fear the calculation for energy is not correct, Please recalculate. (Don't forget to convert rev/s into radians/s)
 
  • #3


is it 86.12HP? I guess I did forget to do that conversion, but is the method correct?
 
  • #4


HI ahello888a,

ahello888a said:

Homework Statement


A rotating uniform cylindrical platform of mass 230 kg and radius 5.6 m slows down from 3.8 rev/s to rest in 16 s when the driving motor is disconnected. Estimate the power output of the motor (\rm hp) required to maintain a steady speed of 3.8 rev/s


Homework Equations


Wfriction=change in KE = -.5*I*w^2
power=work/time


The Attempt at a Solution


So I was was able to get the amount of work that the friction force does and it was -.5 (.5 * MR^2)w^2 = 45053.39J and that work done by friction is the same amount of work needed for the motor to output. So the power of the motor needed is 45053.39/16s = 2690.8Watts and to convert that to HP I just divide by 746W and got 3.6HP. is that right?

I don't believe this is the correct approach. This power that you are finding here is the average power from the frictional torque during the entire slowing down process. But while the speed is decreasing, the power from friction is also decreasing.


So here you need the power at the beginning of the slowing down process. What is the formula for the instantaneous power (for the rotational case)?
 

1. What is the difference between power and torque?

Power is the rate at which work is done, while torque is the measure of rotational force. In simpler terms, power is the speed at which an object can be moved, while torque is the force that causes the object to rotate.

2. Why is horsepower used to measure power and not other units?

Horsepower is a commonly used unit for measuring power because it was historically used to describe the power of steam engines, which were widely used in the past. Today, it is still widely used in the automotive industry to describe the power of engines.

3. How does power and torque affect the performance of a vehicle?

The power and torque of a vehicle are important factors that determine its performance. A higher power output typically results in a faster acceleration and higher top speed, while a higher torque output allows for better towing and hauling capabilities.

4. Can a vehicle have high power and low torque, or vice versa?

Yes, it is possible for a vehicle to have high power and low torque, or vice versa. This largely depends on the design and purpose of the vehicle. For example, a sports car may have high power but low torque, while a truck may have high torque but lower power.

5. How can I increase the power and torque of my vehicle?

There are various ways to increase the power and torque of a vehicle, such as installing a performance exhaust system, upgrading the air intake system, or remapping the engine's computer. However, it is important to note that these modifications may also impact the vehicle's fuel efficiency and reliability.

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