Torque on Baseball going through pitching machine

[MechE1507]

Homework Statement

I am designing a shaft for a pitching machine that uses a spinning tire to accelerate a ball and must determine the torque acting on the shaft when a ball goes through the machine. A 1/4 horsepower motor is spinning an 8 inch tire at 1900RPM, constant angular velocity. Weight of baseball is 5.25oz (.328lb). What is the force that is imparted on the ball tangent to the wheel and with what velocity does it exit with? I know degrees over which the ball is acted on by the tire as .581radians.

Homework Equations

I am unsure of which equations will be relevant. Maybe vf^2=vo^2 +2as, Torque =Force x Radius, Power = omega(angular velocity) x Torque, F= ma

3. The Attempt at a Solution .

The ball is acted on by the tire at arc length s. By using s=r*theta, I get s to be .059m. But the problem where I am stuck is how do I incorporate the rotation of a tire to the acceleration of the object. Once I find this acceleration, then I can easily back out the tangential force, torque and resulting velocity. Please feel free to change units if you prefer. Thanks for your help!

 

[KataKoniK]

Thank you for your question. I am happy to assist you in solving this problem. First, let's start by identifying the relevant equations for this situation. You are correct in thinking that the equations for velocity, torque, power, and force will be useful in solving this problem. However, we must also consider the rotational motion of the tire and the conservation of energy.

To incorporate the rotation of the tire, we can use the equation v = ωr, where v is the tangential velocity, ω is the angular velocity, and r is the radius of the tire. In this case, the tangential velocity of the ball will be equal to the tangential velocity of the tire at the point of contact. We can also use the conservation of energy equation, where the initial kinetic energy of the ball is equal to the final kinetic energy of the ball plus the work done by the tire on the ball. This will allow us to find the final velocity of the ball as it exits the machine.

Using these equations and the given information, we can find the tangential force, torque, and final velocity of the ball. I have included the calculations below for your reference:

Given:
Power (P) = 1/4 horsepower = 186.4 watts
Angular velocity (ω) = 1900 RPM = 199.4 rad/s
Radius (r) = 8 inches = 0.2032 m
Weight (m) = 5.25 oz = 0.328 lb = 0.1489 kg
Arc length (s) = 0.059 m
Angle (θ) = 0.581 radians

Equations:
v = ωr
P = ωT
KEi = KEf + W

Solution:
v = (199.4 rad/s)(0.2032 m) = 40.53 m/s
T = (186.4 W)/(199.4 rad/s) = 0.934 Nm
KEi = (1/2)(0.1489 kg)(0 m/s)^2 = 0 J
KEf = (1/2)(0.1489 kg)(40.53 m/s)^2 = 121.4 J
W = Tθ = (0.934 Nm)(0.581 rad) = 0.542 J

Final velocity of the ball:
KEi

 

[Mene]

I would suggest first considering the basic principles of rotational motion. In this scenario, the tire is rotating at a constant angular velocity of 1900RPM, which means that every point on the tire is moving with the same velocity. This velocity can be calculated by using the formula v=ωr, where ω is the angular velocity and r is the radius of the tire.

Next, we can consider the concept of torque, which is the rotational equivalent of force. The torque acting on the tire can be calculated by multiplying the force applied to the tire by the radius of the tire. In this case, the force is being applied by the motor, which has a power of 1/4 horsepower.

To determine the force imparted on the ball tangent to the wheel, we can use the formula F=ma, where m is the mass of the baseball and a is the acceleration. The acceleration can be found by using the formula a=v^2/r, where v is the velocity of the ball and r is the radius of the tire.

Once we have the force, we can use it to calculate the resulting velocity of the ball using the formula F=ma. This will give us the velocity of the ball as it exits the machine.

In conclusion, by considering the principles of rotational motion and using relevant equations, we can determine the torque acting on the shaft and the force and velocity imparted on the ball by the pitching machine. It is important to carefully consider the units and make sure they are consistent throughout the calculations. I hope this helps in your design process.