The figure shows a ball with mass m = 0.405 kg attached to the end of a thin rod with length L = 0.516 m and negligible mass. The other end of the rod is pivoted so that the ball can move in a vertical circle. The rod is held horizontally as shown and then given enough of a downward push to cause the ball to swing down and around and just reach the vertically up position, with zero speed there.

(a) What initial speed must be given the ball so that it reaches the vertically upward position with zero speed?
(b) What then is its speed at the lowest point?
(c) What then is its speed at the point on the right level with the initial point?
(d) If the ball's mass were doubled, would the answers to (a) through (c) increase, decrease, or remain the same?

Can I start with h = L - L*cos(theta) = L(1 - cos(theta))? Where do I go from there?

(a) Use conservation of energy. To reach the top with zero velocity

(1/2) M Vo^2 = M*g*L
Vo = sqrt(2 g L)

(b) At the lowest point,
(1/2)MV^2 = (1/2) M Vo^2 + M g L
V^2 = Vo^2 + 2 g L
= 4 g L
V = 2 sqrt(g L)

(c) What do you think? The P.E. is the same.

(d) Note that M does not appear in any of the answers above. What does that tell you?

Got it, thanks very much!

Yes, you can start with the equation h = L - L*cos(theta) = L(1 - cos(theta)).

To find the initial speed that must be given to the ball so that it reaches the vertically upward position with zero speed (a), you can use the principle of conservation of mechanical energy. At the highest point, the potential energy is maximum while the kinetic energy is zero. Therefore, the total mechanical energy at the highest point is equal to the initial mechanical energy.

The initial mechanical energy can be calculated as the sum of the potential energy at the highest point (mgh) and the kinetic energy at the start (0.5mv^2).

Substituting h = L(1 - cos(theta)) into mgh and setting it equal to 0.5mv^2, we get:

mgh + 0.5mv^2 = 0

Simplifying the equation, we have:

mgL(1 - cos(theta)) + 0.5mv^2 = 0

Since we want the ball to reach the highest point with zero speed, the final velocity at that point is 0. Therefore, v = 0.

mgL(1 - cos(theta)) + 0.5m(0)^2 = 0

Simplifying further, we get:

mgL(1 - cos(theta)) = 0

Now, solve for the initial speed v.

Next, to find the speed of the ball at the lowest point (b), we can use the principle of conservation of mechanical energy again. At the lowest point, the potential energy is minimum while the kinetic energy is maximum. Therefore, the total mechanical energy at the lowest point is equal to the initial mechanical energy.

The initial mechanical energy can be calculated as the sum of the potential energy at the highest point (mgh) and the kinetic energy at the start (0.5mv^2).

Substituting h = L into mgh and setting it equal to 0.5mv^2, we get:

mgh + 0.5mv^2 = 0

Simplifying the equation, we have:

mgL + 0.5mv^2 = 0

Now, solve for the speed at the lowest point v.

To find the speed at the point on the right level with the initial point (c), we can use the conservation of mechanical energy again. At this point, the potential energy is the same as the initial point but the kinetic energy is different. Therefore, the total mechanical energy at this point is equal to the initial mechanical energy.

The initial mechanical energy can be calculated as the sum of the potential energy at the highest point (mgh) and the kinetic energy at the start (0.5mv^2).

Substituting h = L into mgh and setting it equal to 0.5mv^2, we get:

mgh + 0.5mv^2 = 0

Simplifying the equation, we have:

mgL + 0.5mv^2 = 0

Now, solve for the speed at the point on the right level with the initial point v.

For part (d), if the ball's mass were doubled, the answers to (a) through (c) would remain the same. This is because mass does not affect the total mechanical energy of the system. The total mechanical energy only depends on the height, length, and the gravitational acceleration.

wot m8