Question

A young boy swings a yo-yo horizontally above his head so that the yo-yo has a centripetal acceleration of 200 m/s^2. If the yo-up’s string is 0.33 m long, what is the yo-up’s tangential speed?

Answer 1

v = 8.12 m/s

Step-by-step explanation:

The centripetal acceleration of the yo-yo is given by the following formula:

`a_c = (v^2)/(r)\\\\v^2 = a_cr\\v = √(a_cr)`

where,

v = tangential speed = ?

ac = centripetal acceleration = 200 m/s²

r = radius = length of string = 0.33 m

Therefore, using the given values in the formula, we get:

`v = √((200\ m/^2)(0.33\ m))`

v = 8.12 m/s