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3 years ago

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A small grinding wheel is attached to the shaft of an electric motor which has a rated speed of 3600 rpm. When the power is turn

ed on, the unit reaches its rated speed in 5 s, and when the power is turned off, the unit coasts to rest in 70 s. Assuming uniformly accelerated motion, determine the number of revolutions that the motor executes (a) in reaching its rated speed, (b) in coasting to rest.

1 answer:

dolphi86 [110]3 years ago

6 0

Answer:

(a) θ1 = 942.5rad, (b) θ2 = 13195 rad

Explanation:

(a) Given

ωo = 0 rad/s

ω = 3600rev/min = 3600×2(pi)/60 rad/s

ω = 377rad/s

t1 = 5s

θ1 = (ω + ωo)t/2

θ1 = (377 +0)×5/2

θ1 = 942.5 rads

(b) ωo = 377rad/s

ω = 0 rad/s

t2 = 70s

θ2 = (ω + ωo)t/2

θ2 = (377 +0)×70/2

θ2 = 13195 rad

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According to figure,

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Put the value into the formula

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