# What is the velocity vector of a spinning hoop as view from the ground?

Original Question:

I've calculated a) and b) My issue is with parts c) and d).

I understand that v= 1.56 m/s for the translational motion and that the velocity vector should be the sum of the translational motion vector and the rotational motion vector. However, I'm confused here because the answers to this questions for c is: i. 1.56 + 1.56 = 3.12 ii. 1.56 - 1.56 = 0 iii. $\sqrt{1.56^2 + 1.56^2} My question is, how come they are not doing: i. 1.56 + 2.60 ii. 1.56 - 2.60 iii.$\sqrt{1.56^2 + 2.60^2}\$

Why just use the linear velocity and not the translational velocity as well?

Similarly, for part d), the answer is 1.56 for all of it, but why is it not 2.60 since 2.60 is the translational motion, so if someone were going at the same velocity, they'd be going at 1.56 m/s along with the wheel and should be seeing 2.60 rad/s as the velocity at each point.

You need the translational velocity, you used the translational angular velocity.

$$v = \omega r$$

$$v = 2.60 \times 1.2 \times .5 = 1.56m/s$$

Multiply by .5 to get the radius.

Next time, you can check using the units, because omega has units 1/s while velocity has units m/s.

• But isn't the translational angular velocity equal to the angular velocity, so I might as well always use the angular velocity?
– user40096
Apr 10 '16 at 23:21
• The translational angular velocity is the same as the angular velocity, but different from the translational linear velocity Apr 11 '16 at 4:00