1. Consider a ball on a fixed-length string being whirled in a vertical circular
ID: 1284135 • Letter: 1
Question
1. Consider a ball on a fixed-length string being whirled in a vertical circular path as shown in the diagram below. Use the direction rosette to answer the following questions.
2. You whirl a ball tied to the end of the rope in a horizontal circle at constant speed, as shown in the diagram below. Use the direction rosette to answer the following questions.
(a) What is the direction of the centripetal force acting on the ball when it is at location A?
(b) If the string breaks when the ball is at location A, in what direction will the ball move?
Explanation / Answer
using the angle theta measured with respect to the horizontal , going counterclockwise (as we normally do in trig) we have
Fc (centripetal force) = Tension + component of gravity in direction of tension
component of gravity in direction of tension = mg * sin theta , using a triangle argument is mg sin theta, therefore
Fc = T + mg sin theta
Also keep in mind that centripetal acceleration for constant velocity in a circle is v^2/L
so Fc = mass * acceleration = m*v^2/L
Therefore , an equation that will come useful later:
mv^2/L = T + mg sin theta
a) when ball is at bottom of circle tension is pointing upwards , in the same direction as centripetal force
b) weight always points down and the weight vector is equal to -mg so the answer is 5
c) centripetal force is always center seeking. And it is pointing up here towards the center so 5
d) the centripetal force Fc = T + mg sin theta, here theta = 3pi/2
Fc = T + mg sin(3pi/2)
Fc = T - mg
e) Fc = mv^2/L
f) mv^2 = T -mg
T = mv^2 + mg
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