Rotational torque and kinematics of a rod

In summary, the moment of inertia is calculated to be 1.0 kgm^2, the center of mass is located at (0.6i, 0j), and the magnitude of the gravitational torque is 12.34N*m. After a change in position, the new center of mass is located at (0.56m, -0.2m) and the change in gravitational torque is -4.2N*m. The rotational kinetic energy is equal to 4.1J and the angular velocity is 2.86 rad/s.
  • #1
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Homework Statement
A long thin rod is made out of wood with a uniform density. It has a mass M=2.1 kg and a length L=1.2 meters. The rod is attached to a pivot point at its left end; it is free to rotate around the pivot, but, right now, Fred is holding it horizontal, as shown.

The pivot point sits at the origin of the coordinate system, with the x-axis running to the right and the y-axis pointing upward.

What is the moment of inertia of the rod around the pivot?
Where is the center of mass of this rod? Provide the coordinates in unit-vector notation.

Gravity acts on the rod as if it were pulling directly on the center of mass location. What is the magnitude of the gravitational torque around the pivot point?

Fred now releases the rod. It starts to swing down in response to the gravitational torque. After a short time, the rod has rotated by an angle β=20 degrees.

What is the position of the center of mass now?

What is the change in the gravitational potential energy of the rod, compared to its value when the rod was horizontal? Make sure you supply the appropriate sign.

What is the rotational kinetic energy of the rod at this moment?

What is the angular velocity of the rod at this moment?
Relevant Equations
torque= force * radius* sin(theta) = (moment of inertia) (angular acceleration)
moment of inertia = (1/3) (2.1kg) (1.2m)^2 = 1.0 kgm^2
center of mass= (0.6i, 0j)
magnitude of the gravitational torque=9.8m/s^2*2.1kg*0.6m= 12.34N*m

position of the new center of mass now :
x direction = cos(20)*0.6m=0.56m
y direction= -sin(20) * 0.6m = -0.2m

change in gravitational torque= -(9.8m/s^2*2.1kg*0.6m)sin(20)= -4.2N*m
rotational KE = (1/2) (moment of inertia) (angular velocity)^2
I don't confused on how to proceed further
 

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  • #2
ac7597 said:
change in gravitational torque= -(9.8m/s^2*2.1kg*0.6m)sin(20)= -4.2N*m
Is there a typo here? Did you mean to say "change in gravitational torque?"
rotational KE = (1/2) (moment of inertia) (angular velocity)^2
I don't confused on how to proceed further
Is there any relationship between the change in KE and the change in PE?
 
  • #3
They should be the same in magnitude because conversion of energy.
PE=mass*gravity*height = (2.1kg)(9.8ms/s^2) (-0.2m)= -4.1J
 
  • #4
OK, total energy is conserved. Try to use that.
 
  • #5
thus KE=4.1J
since rotational KE = (1/2) (moment of inertia) (angular velocity)^2
4.1J = (1/2) (1kgm^2) (angular velocity)^2
angular velocity = 2.86 rad/s
 
  • #6
I believe that's correct.
 

1. What is rotational torque?

Rotational torque, also known as torque or moment of force, is a measure of the force that causes an object to rotate around an axis. It is calculated by multiplying the force applied to an object by the distance from the axis of rotation.

2. How is rotational torque different from linear force?

Rotational torque and linear force are both measures of force, but they act on different types of motion. Linear force causes an object to move in a straight line, while rotational torque causes an object to rotate around an axis.

3. What factors affect the rotational torque of a rod?

The rotational torque of a rod is affected by the force applied to it, the distance from the axis of rotation, and the angle at which the force is applied. The material properties and shape of the rod may also play a role.

4. How does rotational torque relate to angular acceleration?

According to Newton's second law of motion, the net torque acting on an object is equal to the moment of inertia of the object times its angular acceleration. This means that the greater the rotational torque, the greater the angular acceleration of the object.

5. What is the formula for calculating rotational torque?

The formula for calculating rotational torque is T = r x F, where T is the torque, r is the distance from the axis of rotation, and F is the force applied to the object. This formula is also known as the cross product of r and F.

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