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A. Compute the torque developed by an industrial motor whose ouputis 150kW at an angular speed of 4000 rev/min.
B. A drum with negligible mass and 0.500m in diameter isattached to the motor shaft, and the power output of the motor isused to raise a weight hanging from a rope wrapped around thedrum. How heavy a weight can the motor lift at constantspeed?
C. At what constant speed will the weight rise?

Respuesta :

lcmendozaf

Answer:

A. Ï„ = 358 N.m

B. m = 73kg

C. V = 209.5 m/s

Explanation:

Let's first convert angular speed to rad/s:

ω = 4000 rev/min * 2π / 60 = 419 rad/s

Since Power is P = τ * ω,

τ = P / ω = 358 N.m

For part B: On the hanging weight:

T - m*g = 0

T = m*g

On the drum:

τ = T*R    Replacing the expression for Tension of the rope:

Ï„ = m*g*R

Solving for m:

m = 73 kg

For part C:

V = ω * R = 209.5 m/s

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