PHysics PHysics Physics Physics Torque

[envscigrl]

Here is my problem:
A 2.50g particle moves in a circle of radius 3.00m. The magnitude of its angular momentum relative to the center of the circle depends on time according to L = (3.5 Nm)t. Find the magnitude of the torque acting on the particle.

I know that:
T = F L
where T = torque
F is the force
L is angular momentum
I feel like this problem is more algebraic than anything. I just need something stable to start from. Pleeeeeeeeaaaase Heeeeeeelp!

 

[jamesrc]

I'm afraid your equation is wrong. You may be thinking of \vec{\tau} = \vec{r}\times\vec{F}, where r is the moment arm (and its a vector equation). What you need to use is the idea that the rate of change of angular momentum is equal to the torque. I hope that helps.

 

[wais]

Hi there,

To solve this problem, we can use the formula T = r x F, where r is the radius of the circle and F is the force acting on the particle. We can also use the formula L = Iω, where I is the moment of inertia and ω is the angular velocity.

First, let's find the angular velocity of the particle. We can rearrange the formula for angular momentum to get ω = L/I. Since the particle is moving in a circle, its moment of inertia can be calculated as I = mr^2, where m is the mass of the particle and r is the radius of the circle. Plugging in the given values, we get I = (2.50g)(3.00m)^2 = 22.5 g m^2.

Now, we can plug in the given values for angular momentum and moment of inertia to find the angular velocity: ω = (3.5 Nm)/(22.5 g m^2) = 0.1556 Nm/g.

Next, we can use the formula for torque to find the force acting on the particle: T = r x F. Rearranging this formula, we get F = T/r. Plugging in the values of torque (3.5 Nm) and radius (3.00m), we get F = 3.5 Nm/3.00m = 1.17 N.

Therefore, the magnitude of the torque acting on the particle is 1.17 N. I hope this helps! Remember to always pay attention to units and use the correct formulas when solving physics problems. Good luck!