Find an expression for the power developed at the given rotation rate

In summary, the conversation discusses the use of 1/N = T to solve for velocity in part (a). The formula F=(m v^2)/R is then used to determine the force of friction, resulting in an answer of 4(π^2)mR(N^2). In part (b), the formula P=Fvcos(theta) is mentioned, but confusion arises due to the angle theta being 90 degrees, resulting in a power of 0. The issue is then addressed and it is clarified that the centripetal force is not of interest and does not equal the force of friction. The focus shifts to discussing the torques exerted on the drum.
  • #1
hi im nimdA
7
0
Homework Statement
A Prony brake is a device that measures the horsepower of engines. The engine rotates a shaft of radius R at N rev/sec. The rotation is opposed by a belt which is attached to two tension measuring devices, such as springs. (a) What is the force of friction acting on the engine? (b) Obtain an expression for the power developed at the given rotation rate.
Relevant Equations
T=(2 pi R)/v , F=(m v^2)/R, P=Fvcos(theta)
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So for part (a), I used the fact that 1/N = the period = T. I solved for the velocity, where i got v=2πRN. I plugged that v into F=(m v^2)/R, which is the centripetal force, but also the force of friction. My answer to part (a) is 4(π^2)mR(N^2).

I'm a bit confused on part (b). I know that P=Fvcos(theta), but since v acts at a tangent and F acts towards the center, my theta will be 90, and
cos 90 = 0, so I'm getting 0 for the power (this doesn't make sense). What am I doing wrong?
 
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  • #2
hi I am nimdA said:
into F=(m v^2)/R, which is the centripetal force, but also the force of friction.
The belt is not moving, and the drum is rigid, so I'm not sure what mass your m is, but centripetal force cannot be of interest. And it certainly does not equal the force of friction.
What torques are being exerted on the drum?
 

Related to Find an expression for the power developed at the given rotation rate

1. What does it mean to find an expression for power developed at a given rotation rate?

When we talk about finding an expression for power developed at a given rotation rate, it means we are trying to determine the mathematical relationship between power and rotation rate. In other words, we want to find an equation that can be used to calculate the power developed at a specific rotation rate.

2. How is power related to rotation rate?

Power and rotation rate are directly proportional to each other. This means that as the rotation rate increases, the power developed also increases. Similarly, when the rotation rate decreases, the power developed also decreases.

3. What are the units of power and rotation rate?

The standard unit for power is watts (W) and the standard unit for rotation rate is revolutions per minute (RPM). However, other units such as horsepower (hp) and radians per second (rad/s) may also be used for power and rotation rate, respectively.

4. Can you provide an example of an expression for power developed at a given rotation rate?

Yes, an example of an expression for power developed at a given rotation rate is P = 2πNT, where P is power in watts, N is the rotation rate in RPM, and T is the torque in Newton-meters (Nm).

5. How is the expression for power developed at a given rotation rate derived?

The expression for power developed at a given rotation rate is derived using the formula P = τω, where P is power, τ is torque, and ω is angular velocity. By rearranging this formula and converting angular velocity from radians per second to revolutions per minute, we can derive the expression P = 2πNT.

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