Two Conceptual questions about rotational mechanics

In summary, in physics problems where the cord on a pulley does not slip, friction can be ignored. In the given problem, the assumption that the cord will have the same final speed throughout the system is valid because the tension from the mass and the moment of inertia of the sphere will keep the rope from stretching. This allows for the use of the formula for the sum of kinetic energy being equal to the initial potential energy.
  • #1
babayevdavid
17
0
Hi,

I see in many of these physics problems it is explicitly stated that "the cord on the pulley does not slip." [1] What is the implication of this? Why is it important?

Also, there is a problem here with a diagram that I will try to explain in words and then have a question about. (Assuming frictionless axes) There is a sphere rotating around a vertical axis through its center. Around its equator is tied some massless cord that continues horizontally from the sphere onto a pulley (I and radius of the sphere and pulley are given). The cord hangs over the pulley and at the end of the cord is a mass. If the mass were to be dropped a certain distance vertically from rest, and I wanted to find its final speed, and I used the sum of all kinetic energy present = the initial potential energy of the mass at rest, [2] is it okay to substitute the omegas in that setup for V/r ? Why is that okay? It is a valid assumption that the cord will have the same final speed throughout the system?

Thanks in advance!

-David
 
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  • #2
The bit about the rope not slipping is usually so that you can ignore friction.

It is a valid assumption that the cord will have the same final speed throughout the system

Yes. You have two things keeping it in tension, the mass pulling in one direction and the moment of inertial of the sphere "pulling" in the other. If it doesn't stretch the rope will all be going at the same speed.
 

1. How does rotational motion differ from linear motion?

Rotational motion is the movement of an object around an axis, while linear motion is the movement of an object in a straight line. In rotational motion, the object's orientation changes, while in linear motion, the object's position changes.

2. What is the relationship between torque and angular acceleration?

Torque is the rotational equivalent of force, and it is the product of force and the distance from the fulcrum. Angular acceleration is the rate of change of angular velocity. The relationship between torque and angular acceleration is given by the equation: torque = moment of inertia * angular acceleration. This means that for a given torque, a smaller moment of inertia will result in a larger angular acceleration.

3. How do you calculate the moment of inertia for a rigid body?

The moment of inertia for a rigid body can be calculated by summing up the individual moments of inertia for each small element of the body, using the formula: I = ∑m*r^2, where m is the mass of the element and r is the distance from the axis of rotation. The moment of inertia depends on the shape and distribution of mass in the object.

4. Can you have rotational motion without a torque?

No, rotational motion cannot occur without a torque. This is because torque is the force that causes rotational motion. Without an external torque acting on an object, it will continue to remain in its state of rest or uniform motion.

5. How does the center of mass affect rotational motion?

The center of mass is the point at which the entire mass of an object can be considered to be concentrated. It plays a crucial role in rotational motion, as it is the point about which an object rotates. The center of mass also determines the stability of an object, with a lower center of mass resulting in greater stability.

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