Question

What is the magnitude of the tangential acceleration of a bug on the rim of a 13.0-in.-diameter disk if the disk accelerates uniformly from rest to an angular speed of 79.0 rev/min in 3.80 s

Answer 1

The correct solution will be "12.48 in/s²".

Step-by-step explanation:

The given values are:

Angular speed

= 79.0 rev/min

Time

= 3.80 seconds

Now,

⇒
`Angular \ acc.=(((79* 2* \pi)/(60)))/(3.8)`

`=(8.2)/(3.8)`

`=2.1 \ rad/s^2`

Tangential acceleration will be:

=
`Angular acc.* radius`

=
`2.1* 6.5* 0.0254`

=
`0.3467 \ m/s^2`

=
`12.48 \ in/s^2`