Can Rotation Affect the Linear Motion of an Object's Center of Mass?

AI Thread Summary
The discussion revolves around determining the initial velocity of a spinning ball moving upwards under gravitational force. The participant found differing results using mechanical energy and kinematics, suspecting that the discrepancy arises from including rotational kinetic energy in the calculations. It was clarified that rotational motion does not affect the linear motion of the center of mass, and additional energy from rotation should not be included in vertical motion equations. Ultimately, the participant realized that even with rotational energy considered, the terms cancel out when the center of mass velocity is zero. This insight resolves the confusion regarding the differing initial velocity calculations.
dieyn
Messages
2
Reaction score
0

Homework Statement


This isn't so much of a problem as a general question. I am trying to find the starting velocity of a spinning ball going upwards (in air, close to Earth's surface, only force acting on it is the gravitational force) until its linear velocity reaches zero. I found the initial velocity two different ways (with mechanical energy and with kinematics), and I am getting answers that differ by a constant.

Homework Equations


mgh=\frac{7}{10}mv_0^2

v_f^2=v_0^2-2gh
(where the final velocity is 0)

The Attempt at a Solution


When I try to find the initial velocity, I can see that the two differ by a constant. I know that if the ball had no rotational kinetic energy, the equations would line up. However, I thought that the rotational motion would not have an effect on the linear motion of its center of mass. I think that the discrepancy is because the kinematics equation is derived from the conservation of energy of an object that has no rotational motion, but I'm not sure. I appreciate your help!
 
Physics news on Phys.org
Yes. If you set mgh = \frac{1}{2}mv2 then you get that the initial velocity is \sqrt{2gh}. This is the same as the result from the kinematics equation when the final velocity is zero.

Any additional energy that you add on top of the KE will cause the answers to be different. There's no need to include the rotational energy in a treatment of the vertical motion. For the same reasons that you wouldn't include, for example, the internal energy of the particles that make up the ball.
 
Oh, wait. I think I just realized my error. When the ball reaches the point where its center of mass has a velocity of zero, the ball still retains its initial angular velocity. Therefore, the terms cancel out in the end, even if you include the rotational kinetic energy in the equation. Thank you for your help!
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top