Question

A 0.160-kg ball attached to a light cord is swung in a vertical circle of radius 70.0 cm. At the top of the swing, the speed of the ball is 3.26 m/s. The centre of the circle is 1.50 m above the floor.
Calculate the speed of the ball when the cord is 30.0° down from horizontal.

Answer 1

@dan815

Answer 2

I drew a free body diagram, but I'm stuck from there

Answer 3

draw out the question it will help. at t=0 v initial is at 3.26 m/s. calculate the components of the velocity at the angle 30 degrees from the horizontal. F(circular) = m* v^2/R you also have force of gravity F=mg.

Answer 4

Yeah, I drew it out but I'm not sure how to calculate the components

Answer 5

if you find the force at that point you can solve for velocity as long as you are using net force

Answer 6

you know gravity only effects the y component of the force and the F(circular) has both x and y

Answer 7

Yeah

Answer 8

so first find the inital conidions. so the total net force at the beginning

Answer 9

conditions*

Answer 10

The force of tension I found to be 0.86 N, and centripetal I found to be 1.7 N. (For the top of the circle)

Answer 11

right so tension at the top is only in the x direction
at the beginning and the fg is in the y

Answer 12

The x-direction? I thought it was acting vertically at the top

Answer 13

\[F_t\] at the horizontal is pulling the object so it doesn't escape. there is also a force of Gravity \[F_g\] and a force due to the inital velocity

Answer 14

I see

Answer 15

to make sure is the top of the swing meaning at this point