Find Torque given Angular momentum and time

[btlogan2]

Homework Statement

A particle is to move in an xy plane, clockwise around the origin as seen from the positive side of the z axis. In unit-vector notation, what torque acts on the particle at time t = 5.3 s if the magnitude of its angular momentum about the origin is (a) 5.7 kg·m2/s, (b) 5.7t2 kg·m2/s3, (c) 5.7t1/2 kg·m2/s3/2, and (d) 5.7/t2 kg·m2*s?

Additionally, the answers have to be in terms of i, j, and k.

Homework Equations

Torque= r X F

L= r x p = m(r x v)

Tnet = dL/dt

The Attempt at a Solution

I don't understand how to use what is given into any of the formulas. Torque is equal to the time derivative of the angular momentum. So how would I go about solving this.

 

[vela]

Homework Statement

A particle is to move in an xy plane, clockwise around the origin as seen from the positive side of the z axis. In unit-vector notation, what torque acts on the particle at time t = 5.3 s if the magnitude of its angular momentum about the origin is (a) 5.7 kg·m2/s, (b) 5.7t2 kg·m2/s3, (c) 5.7t1/2 kg·m2/s3/2, and (d) 5.7/t2 kg·m2*s?

Additionally, the answers have to be in terms of i, j, and k.

Homework Equations

Torque= r X F

L= r x p = m(r x v)

Tnet = dL/dt

The Attempt at a Solution

I don't understand how to use what is given into any of the formulas. Torque is equal to the time derivative of the angular momentum. So how would I go about solving this.

Like you said, the torque is the time derivative of the angular momentum, so differentiate the angular momentum. First, you need to understand the angular momentum of the object. You're given the magnitude of the angular momentum as a function of time. What about its direction?

 

[btlogan2]

How do you find the direction with the information provided?

 

[vela]

The direction of the angular momentum is directed along the rotational axis, according to the right-hand rule. The first sentence provides the needed info.

 

[Nickie]

I would approach this problem by first understanding the given information and equations. From the information provided, we know that the particle is moving in a clockwise direction in the xy plane and that we need to find the torque acting on it at a specific time. We also have four different expressions for the magnitude of the angular momentum at that time, each with a different time dependence.

Using the equations for torque and angular momentum, we can set up an equation to solve for the torque. We know that torque is equal to the time derivative of angular momentum, so we can use the equation Tnet = dL/dt. This means that the torque is equal to the change in angular momentum over time.

For part (a), the magnitude of the angular momentum is given as a constant value of 5.7 kg·m2/s. Therefore, the time derivative of this value is 0, and the torque would be 0.

For part (b), the magnitude of the angular momentum is given as 5.7t2 kg·m2/s3. We can take the time derivative of this expression to get the torque, which would be 11.4t kg·m2/s2.

For part (c), the magnitude of the angular momentum is given as 5.7t1/2 kg·m2/s3/2. Again, we can take the time derivative of this expression to get the torque, which would be 2.85t-1/2 kg·m2/s3/2.

For part (d), the magnitude of the angular momentum is given as 5.7/t2 kg·m2*s. We can take the time derivative of this expression to get the torque, which would be -11.4/t3 kg·m2/s2.

To express these torque values in terms of i, j, and k, we would need to use the cross product formula T = r x F and the fact that the torque vector is perpendicular to both the position vector (r) and the force vector (F). Without more information about the force acting on the particle, we cannot determine the exact values for the torque vector in unit-vector notation. However, we can use the calculated values above as the magnitudes of the torque vector and assign arbitrary directions for each component (i, j, and k) to represent the direction of the torque vector.