Question

A .275 kg. object is swung in a vertical circular path on a string .850 m. long. A) What are the forces acting on the ball at any point along this path? B) Draw free-body diagrams for the ball when it is at the bottom of the circle and when it is at the top. C) If its speed is 5.2m/s at the top of the circle, what is the tension in the string there? D) If the string breaks when its tension exceeds 22.5 N. what is the maximum speed the object can have at the bottom before the string breaks?

Answer 1

m g down
Tension of string toward center
At top
m g + T = m v^2/r
T = m (v^2/r-9.81
T = .275(5.2^2/.85 -9.81)
= 6.05 N

At bottom
T - m g = m v^2/r
22.5 = .275 (v^2/.85 + 9.81)

Answer 2

c) F = ma (in the 'y' direction)

*ball at top means 'T' and 'mg' are pointing "downward"*

-T -mg = ma
-T = mg + ma
T = -mg -ma
T = -m(g + v^2/r)

T = -.275(9.8 + (5.20^2/.85))

T = -11.4 N, or, 11.4 N downward

Answer 3

d) F = ma (in the 'y' direction)

*ball at bottom mean 'T' is upward and 'mg' is downward*

T - mg = ma
T - mg = m(v^2/r)
(r(T - mg))/m = v^2
v = sqrt((r(T - mg))/m)

v = sqrt((.85(22.5 - (.275*9.8))/.275)

v = 7.82 m/s

Answer 4

need a fee body diagram