Rod rotates when hit by point mass/ Cons. of angular momentum

In summary, a 4m long stick with a mass of 3 kg is hung vertically and a point mass of 7 kg impacts the bottom-most point while traveling at 6.11 m/s. Using the conservation of angular momentum theorem, the angular speed after the collision is found to be 1.38 rad/sec. The rotational kinetic energy of the system after the collision is used to determine the change in height of the center of mass, which is then converted to an angle of 67.92 degrees. The moment of inertia of the stick was not calculated correctly and needs to be redone.
  • #1
Epictetus
5
0
A thin uniform stick of length 4m and mass 3 kg is hung vertically from a support 1m from its end. The stick is initially at rest. A point mass of 7 kg traveling at 6.11 m/s is moving and impacts the stick at the bottom-most point. Determine how high the stick and point mass will rotate before coming to rest momentarily.

I applied the conservation of angular momentum theorum

L = r (moment arm) mv
= (3)(7)(6.11) = 128.31

I = (summation) m r(moment arm)^2 = 3 (1)^2 + 3 (3)^2 = 30 kg m^2
(where 1 and 3 are the distances from the pivot point to the center of mass and pivot point to the point of impact, respectively)

I` = I(rod) + 7(3)^2 = 93 Kg m^2

L = I omega
128.31 = 93 omega
omega = 1.38 rad/sec

Kinetic energy is then transferred to potenial:

1/2 I` omega ` ^2 = mgh
1/2 (93) (1.38) ^2 = mgh
88.51 = g[7 (3cos theta) + 3 cos theta]
9.022 = 24 cos theta
theta = 67.92 deg.

Are the moment arms that I used for the (I = (summation) m r(moment arm)^2) formula correct? Assuming I did my calculations correct, how do I determine the height from the information that I got?
 
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  • #2
Epictetus said:
A thin uniform stick of length 4m and mass 3 kg is hung vertically from a support 1m from its end. The stick is initially at rest. A point mass of 7 kg traveling at 6.11 m/s is moving and impacts the stick at the bottom-most point. Determine how high the stick and point mass will rotate before coming to rest momentarily.
I assume that the point mass attaches itself to the stick? (An inelastic collision?)

I applied the conservation of angular momentum theorum

L = r (moment arm) mv
= (3)(7)(6.11) = 128.31
OK.

I = (summation) m r(moment arm)^2 = 3 (1)^2 + 3 (3)^2 = 30 kg m^2
(where 1 and 3 are the distances from the pivot point to the center of mass and pivot point to the point of impact, respectively)
I assume this is supposed to be the moment of inertia of the stick? If so, you'll have to redo it. What's the moment of inertia of a stick about one end? or about its center? (Look it up!) Then use the parallel axis theorem (Or just treat it as two short sticks added together) to find the moment of inertia about the pivot point.

...

Are the moment arms that I used for the (I = (summation) m r(moment arm)^2) formula correct?
No. See comments above.
Assuming I did my calculations correct, how do I determine the height from the information that I got?
Use conservation of angular momentum to find the angular speed after the collision. Then find the rotational KE of the system after the collision. Then apply conservation of mechanical energy. (Hint: Find the change in height of the center of mass, then convert that to an angle.)
 
  • #3


Yes, the moment arms you used for the moment of inertia formula are correct. To determine the height, you can use the equation for potential energy: mgh = 1/2 I` omega`^2, where h is the height and omega` is the angular velocity at the moment of impact. Rearranging for h, you get h = (1/2 I` omega`^2)/mg. Plugging in the values you calculated, you get h = (1/2 * 93 * 1.38^2)/9.8 = 8.28 meters. This is the maximum height that the stick and point mass will rotate before coming to rest momentarily.
 

What is the principle of conservation of angular momentum?

The principle of conservation of angular momentum states that the total angular momentum of a system remains constant unless an external torque is applied. This means that angular momentum cannot be created or destroyed, only transferred between objects in a system.

How does a rod rotate when hit by a point mass?

When a point mass hits a rod, it exerts an external torque on the rod. According to the principle of conservation of angular momentum, the total angular momentum of the system must remain constant. This causes the rod to rotate in the opposite direction to the incoming point mass, in order to maintain the system's total angular momentum.

What factors affect the rotation of a rod when hit by a point mass?

The rotation of a rod when hit by a point mass is affected by the momentum of the point mass, the distance between the point of impact and the rod's axis of rotation, and the mass and shape of the rod itself. These factors can determine the speed, direction, and extent of the rod's rotation.

Is the principle of conservation of angular momentum always applicable?

Yes, the principle of conservation of angular momentum is a fundamental law of physics that applies to all isolated systems. In other words, as long as no external torques are acting on a system, the total angular momentum of that system will remain constant.

How is the principle of conservation of angular momentum used in real-world applications?

The principle of conservation of angular momentum is used in various real-world applications, such as in spacecraft navigation, gyroscopes, and sports equipment like ice skates and bicycles. It also plays a crucial role in understanding the dynamics of rotating systems, such as planets and galaxies.

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