Question

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 motor 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%?

Answer 1

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 %.

Step-by-step 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[(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[(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* 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* 10^(-2)\,m)\cdot (-10)\cdot (1.5\,s)`

`a_(t) = -1.075\, (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\,(rad)/(s)` and
`\omega = 26.180\,(rad)/(s)`, then the time needed is:

`26.180\,(rad)/(s) = 104.720\,(rad)/(s)-5\cdot t^(2)`

`5\cdot t^(2) = 104.720\,(rad)/(s)-26.180\,(rad)/(s)`

`t^(2) = (104.720\,(rad)/(s)-26.180\,(rad)/(s) )/(5)`

`t = \sqrt{(104.720\,(rad)/(s)-26.180\,(rad)/(s) )/(5) }`

`t \approx 3.963\,s`

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