Angular momentum and torque of a particle

[physics_geek]

Homework Statement

A particle is located at the vector position =(4.00i + 5.00j) m and a force exerted on it is given by =(2.00i + 1.00j) N.
a. What is the torque acting on the particle about the origin?
b. Consider another point about which the torque caused by this force on this particle will be in the opposite direction and half as large in magnitude. Select the following conditions that are true.
No such point can exist.
Only one such point can exist.
Multiple such points can exist.
No such a point can lie on the y-axis.
Only one such point can lie on the y-axis.
Multiple such points can lie on the y-axis.

c. Determine the position vector of such a point.

Homework Equations

cross products A x B

The Attempt at a Solution

so i got part a..its -6k
i need help with parts b and c please

 

[jeffreydk]

First, you know that a force in the opposite direction and half the magnitude is just $3\hat{k}$. So take a look at the torque;

$$\tau=\mathbf{r}\times\mathbf{F}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a & b & 0 \\ 2 & 1 & 0 \end{vmatrix} =[a-2b]\hat{k}$$

So what possiblities could you have here for $a$ and $b$ (which are of course the position components) if you know the torque must be equal to $3\hat{k}$?

 

[nlightner]

I would like to clarify that angular momentum and torque are fundamental concepts in physics that describe the rotational motion of a particle or a system of particles. Angular momentum is the measure of the rotational inertia of an object, while torque is the measure of the rotational force acting on an object. Both of these quantities are vector quantities and can be calculated using the cross product of the position vector and the force vector.

Now, let's move on to the specific problem given. The particle is located at the position vector (4.00i + 5.00j) m, and a force of (2.00i + 1.00j) N is acting on it. Using the cross product formula, we can calculate the torque acting on the particle about the origin as follows:

τ = r x F
= (4.00i + 5.00j) x (2.00i + 1.00j)
= 4.00(1.00j) - 5.00(2.00i)
= -8.00i + 4.00j
= -8.00k

Therefore, the torque acting on the particle about the origin is -8.00k.

Moving on to part b, we are asked to consider another point about which the torque caused by this force on the particle will be in the opposite direction and half as large in magnitude. From the formula for torque, we can see that the direction of torque is determined by the cross product of the position vector and the force vector. So, for the torque to be in the opposite direction, the position vector and the force vector must be anti-parallel, i.e. they must have opposite directions.

This means that the position vector must be equal to the force vector multiplied by a constant factor of -2. So, the position vector can be written as (-4.00i - 2.00j) m. This satisfies the condition for the torque to be in the opposite direction and half as large in magnitude.

Now, let's consider the point lying on the y-axis. The y-axis is perpendicular to both the position vector and the force vector, so the cross product of these two vectors will be parallel to the y-axis. From the formula for torque, we can see that the magnitude of torque is determined by the product of the magnitudes of the position vector and the force vector.