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frosja888 [35]
3 years ago
12

An electric motor spins at 1000 rpm and is slowing down at a rate of 10 t rad/s2 ; where t is measured in seconds. (a) If the mo

tor radius is 7.165 cm, what is the tangential component of acceleration at the edge of the motor at t = 1.5 seconds? (b) How long will it take, in seconds, to decrease its angular velocity by 75%?
Physics
1 answer:
REY [17]3 years ago
8 0

Answer:

a) The tangential component of acceleration at the edge of the motor at  t = 1.5\,s is -1.075 meters per square second.

b) The electric motor will take approximately 3.963 seconds to decrease its angular velocity by 75 %.

Explanation:

The angular aceleration of the electric motor (\alpha), measured in radians per square second, as a function of time (t), measured in seconds, is determined by the following formula:

\alpha = -10\cdot t\,\left[\frac{rad}{s^{2}} \right] (1)

The function for the angular velocity of the electric motor (\omega), measured in radians per second, is found by integration:

\omega = \omega_{o} - 5\cdot t^{2}\,\left[\frac{rad}{s} \right] (2)

Where \omega_{o} is the initial angular velocity, measured in radians per second.

a) The tangential component of aceleration (a_{t}), measured in meters per square second, is defined by the following formula:

a_{t} = R\cdot \alpha (3)

Where R is the radius of the electric motor, measured in meters.

If we know that R = 7.165\times 10^{-2}\,m, \alpha = 10\cdot t and t = 1.5\,s, then the tangential component of the acceleration at the edge of the motor is:

a_{t} = (7.165\times 10^{-2}\,m)\cdot (-10)\cdot (1.5\,s)

a_{t} = -1.075\, \frac{m}{s^{2}}

The tangential component of acceleration at the edge of the motor at  t = 1.5\,s is -1.075 meters per square second.

b) If we know that \omega_{o} = 104.720\,\frac{rad}{s} and \omega = 26.180\,\frac{rad}{s}, then the time needed is:

26.180\,\frac{rad}{s} = 104.720\,\frac{rad}{s}-5\cdot t^{2}

5\cdot t^{2} = 104.720\,\frac{rad}{s}-26.180\,\frac{rad}{s}

t^{2} = \frac{104.720\,\frac{rad}{s}-26.180\,\frac{rad}{s}  }{5}

t = \sqrt{\frac{104.720\,\frac{rad}{s}-26.180\,\frac{rad}{s}  }{5} }

t \approx 3.963\,s

The electric motor will take approximately 3.963 seconds to decrease its angular velocity by 75 %.

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