Energy and momentum - pendulums

[FrenchAtticus]

Homework Statement

A heavy ball swings on a string in a circular arc of radius 1m.
The two highest points of the ball's trajectory are Q and Q'; at these points the string is +/- 20 degrees from the vertical. Point P is the lowest point of the ball's trajectory where the string hangs vertically down. The acceleration of gravity is 9.8 m/s^2.

1.) What is the ball's speed at the point P? Neglect air resistance and other frictional forces.
2.) What is the magnitude of the ball's acceleration at the point P?
3.) What is the ball's speed at the point Q?
4.) What is the magnitude of the ball's acceleration at the point Q?

Homework Equations

mgh1 + 1/2mv1^2 = mgh2 + 1/2mv2^2

The Attempt at a Solution

9.8(1-cos(20)) + 1/2v1^2 = 0 + 1/2v2^2
0.591 + 1/2v1^2 = 1/2v2^2

How can I solve this with two unknowns?

 

[LowlyPion]

Why do you think there are 2 velocities to consider?

PE at highest = KE at lowest

m*g*h = m*v²/2

 

[thomate1]

1.) To find the ball's speed at point P, we can use the conservation of energy equation: mgh1 + 1/2mv1^2 = mgh2 + 1/2mv2^2. Since the ball is at its lowest point at P, h1 = 0 and h2 = 1m. We also know that the ball's velocity at Q is perpendicular to the string, so v1 = 0. Solving for v2, we get v2 = √(2gh2) = √(2*9.8*1) ≈ 4.43 m/s. This is the ball's speed at point P.

2.) To find the magnitude of the ball's acceleration at point P, we can use the equation a = v^2/r, where v is the ball's speed and r is the radius of the circular arc. Since the ball is at its lowest point at P, the radius is equal to the length of the string, which is 1m. Substituting v = 4.43 m/s and r = 1m, we get a = (4.43)^2/1 = 19.6 m/s^2. This is the magnitude of the ball's acceleration at point P.

3.) To find the ball's speed at point Q, we can use the same conservation of energy equation as in part 1. This time, h1 = h2 = 1m since the ball is at the same height at both points Q and Q'. Solving for v2, we get v2 = √(2gh2) = √(2*9.8*1) ≈ 4.43 m/s. This is the ball's speed at point Q.

4.) To find the magnitude of the ball's acceleration at point Q, we can again use the equation a = v^2/r. However, since the ball is at an angle of 20 degrees at point Q, the radius of the circular arc is now the length of the string multiplied by the cosine of 20 degrees. So, r = 1*cos(20) ≈ 0.94m. Substituting v = 4.43 m/s and r = 0.94m, we get a = (4.43)^2/0.94 ≈ 20.8 m/s^