Question

A small grinding wheel is attached to the shaft of an electric motor which has a rated speed of 4200 rpm. When the power is turned 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. Assume uniformly accelerated motion. determine the number of revolutions that the motor executes

(a) in reaching its rated speed,
(b) in coating to rest.

Answer 1

a)
`\ddot n = 50400\,(rev)/(min^(2))`, b)
`n = 2450\,rev`

Step-by-step explanation:

a) The acceleration experimented by the grinding wheel is:

`\ddot n = (4200\,(rev)/(min) - 0 \,(rev)/(min) )/((5)/(60)\,min )`

`\ddot n = 50400\,(rev)/(min^(2))`

Now, the number of revolutions done by the grinding wheel in that period of time is:

`n = ((4200\,(rev)/(min) )^(2)-(0\,(rev)/(min) )^(2))/(2\cdot \left(50400\,(rev)/(min^(2)) \right))`

`n = 175\,rev`

b) The acceleration experimented by the grinding wheel is:

`\ddot n = (0 \,(rev)/(min) - 4200\,(rev)/(min) )/((70)/(60)\,min )`

`\ddot n = -3600\,(rev)/(min^(2))`

Now, the number of revolutions done by the grinding wheel in that period of time is:

`n = ((0\,(rev)/(min) )^(2) - (4200\,(rev)/(min) )^(2))/(2\cdot \left(-3600\,(rev)/(min^(2)) \right))`

`n = 2450\,rev`