Question

The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 7.0 rev/s in 7.0 s. At this point, the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub slows to rest in 13.0 s. Through how many revolutions does the tub turn during this 20 s interval? Assume constant angular acceleration while it is starting and stopping.

Answer 1

70 revolutions

Step-by-step explanation:

t = Time taken

`\omega_f` = Final angular velocity

`\omega_i` = Initial angular velocity

`\alpha` = Angular acceleration

`\theta` = Number of rotation

Equation of rotational motion

`\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=(\omega_f-\omega_i)/(t)\\\Rightarrow \alpha=(7-0)/(7)\\\Rightarrow a=1\ rev/s^2`

`\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=(\omega_f^2-\omega_i^2)/(2\alpha)\\\Rightarrow \theta=(7^2-0^2)/(2* 1)\\\Rightarrow \theta=24.5\ rev`

Number of revolutions in the 7 seconds is 24.5

`\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=(\omega_f-\omega_i)/(t)\\\Rightarrow \alpha=(0-7)/(13)\\\Rightarrow a=-0.5384\ rev/s^2`

`\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=(\omega_f^2-\omega_i^2)/(2\alpha)\\\Rightarrow \theta=(0^2-7^2)/(2* -0.5384)\\\Rightarrow \theta=45.5\ rev`

Number of revolutions in the 13 seconds is 45.5

Total total number of revolutions in the 20 second interval is 24.5+45.5 = 70 revolutions