A pitcher hurls a 0.25kg ball around a vertical circular path of radius 0.6 m, applying a tangential force of 30N, before releasing it at the bottom of the circle (underhand pitch). If the speed of the ball at the top of the circle was been 15 m/s, what will be the speed just after it's released?
I need some help! Please show the equation and then plug in the numbers into that equation
Why the 1.2?
Ke up top = 1/2*.25 *225
gain of Pot energy = .25*9.81*1.2
work input = (1/4)(2 pi *.6)*30
so
sum of those 3 energies =
(1/2)(.25)v^2
Thank you so much!
To find the speed of the ball just after it's released, we can use the principle of conservation of mechanical energy. At the top of the circular path, the ball has both kinetic energy and potential energy. Just after it's released at the bottom of the circle, it will have only kinetic energy.
First, let's find the potential energy of the ball at the top of the circle. The potential energy is given by the equation:
PE = mgh
where m is the mass of the ball, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height of the ball above the bottom of the circle. In this case, h = 2r, where r is the radius of the circle. Thus, h = 2(0.6 m) = 1.2 m.
Next, let's find the potential energy of the ball at the top:
PE = (0.25 kg)(9.8 m/s²)(1.2 m)
Now, let's find the initial kinetic energy of the ball at the top. The kinetic energy is given by:
KE = 0.5mv²
where v is the speed of the ball. In this case, the given speed is 15 m/s.
KE = 0.5(0.25 kg)(15 m/s)²
Now, since the mechanical energy is conserved, the total energy at the top (potential energy plus kinetic energy) will be the same as the total energy just after the release (kinetic energy).
PE + KE = KE'
Substituting the values we found:
(0.25 kg)(9.8 m/s²)(1.2 m) + 0.5(0.25 kg)(15 m/s)² = 0.5(0.25 kg)v'²
Simplifying the equation:
2.94 J + 56.25 J = 0.125 Jv'²
Now, let's solve for v'²:
59.19 J = 0.125 Jv'²
Divide both sides of the equation by 0.125 J:
v'² = 59.19 J / 0.125 J
v'² = 473.52
Taking the square root of both sides of the equation:
v' = √473.52
v' ≈ 21.76 m/s
Therefore, the speed of the ball just after it's released will be approximately 21.76 m/s.