Will a Rigid Body Continue to Rotate Without Centripetal Force?

[player1_1_1]

Homework Statement

hello, I need help with projection of a rigid body with moment of inertia $$I$$, the rigid body was earlier moving on a circle of $$R$$ radius with $$\omega$$ angular velocity and was making angle of $$\alpha$$ when centripetal force stopped to work. And I need to know if this rigid body will have angular velocity when moving?

 

[Kaleb]

Think about this,

The Earth has moment of inertia I, it is moving around the Sun at radius R and angular velocity w. It is currently at some angle w.r.t. the Sun. Now let's take the Sun (and the rest of the mass except for the Earth) out of the picture. Will the Earth continue to rotate around a Starless solar system after the Sun disappeared? Will it have linear velocity?

-MarkP.S. Grab a tether ball and swing it around your head then let the string go.

 

[gneill]

The question is a bit vague on details. Certainly after the object "has its string cut" and it heads off linearly, it will still have angular velocity with respect to the old center of motion. But it will no longer be in circular motion, it's just the rate of change of the angle of line-of-sight from the center to the departing object.

Now, if the question was referring to the intrinsic rotation of the object about its own axis, then that will depend upon the details of its motion prior to release. Was it always presenting the same face to the center?

 

[player1_1_1]

thanks for answers, yeah the same face to center, so it will rotate with same velocity that it had earlier but move on straight line?

 

[gneill]

thanks for answers, yeah the same face to center, so it will rotate with same velocity that it had earlier but move on straight line?

Why don't you examine the angular momenta of the system? Before being set free the body was orbiting around a center. It was also rotating (it must have been if it always had one face towards the center).

Can you account for all of the angular momentum before and after the split?

 

[player1_1_1]

so if it was rotating with $$\omega_1$$ radial velocity, from conservation of energy will be
$$I\omega_1^2=Mv^2+I\omega_2^2$$ where $$\omega_1$$ is start and $$\omega_2$$ is finish velocity, now we know that $$v=\omega d$$ so it will be $$I\omega_2^2=I\omega_1^2-Md^2\omega_1^2$$ and its easy now to find $$\omega_2$$, good?

 

[gneill]

Here are my thoughts.

Have you considered using the parallel axis theorem to find the angular momentum before the split? After all, the object is essentially rotating about an eccentric axis if it always has one face towards that center.

L = (I + m*R^2)*w

When the split happens, the angular momentum that was in the orbit can be identified with the term m*R^2*w, since at the instant of the split V = w*R and the angular momentum for the linear trajectory will be m*(r x V )= m*R*(w*R )= m*w^2*R.

That leaves the I*w term untouched. So the body should be rotating about its own center of gravity with an angular velocity w.

 

[player1_1_1]

thx;)