Consider a ball on a fixed-length string being whirled in a vertical circular pa

Question

Consider a ball on a fixed-length string being whirled in a vertical circular path as shown in the diagram below.

(a) When the ball is at the bottom of the circle, what is the direction of the tension force in the string? Use the direction rosette to answer this question.
---Select--- 1 2 3 4 5 6 7 8 9

(b) When the ball is at the bottom of the circle, what is the direction of the gravitational force on the ball? Use the direction rosette to answer this question.
---Select--- 1 2 3 4 5 6 7 8 9

(c) When the ball is at the bottom of the circle, what is the direction of the centripetal force on the ball? Use the direction rosette to answer this question.
---Select--- 1 2 3 4 5 6 7 8 9

(d) When the ball is at the bottom of the circle, what is the expression for the centripetal force on the ball? Take the upward direction as positive and downward direction as negative when considering the sign of the forces. (Use the following as necessary: m, g, and T.)
Fc =   

(e) If the speed of the ball at the bottom of the circle is v, what is the expression for the centripetal force on the ball in terms of its speed v and radius L? Take the upward direction as positive and downward direction as negative when considering the sign of the forces. (Use the following as necessary: m, v, and L.)
Fc =   

(f) Use your answers in parts (d) and (e) to get an expression for the tension in the string in terms of the speed of the ball and the length of the string. (Use the following as necessary: m, v, L, and g.)
T =  

Explanation / Answer

Using the angle theta measured with respect to the horizontal , going counterclockwise gravity in direction of tension 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
ANSWER----
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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