a 8.20 kg particle P has a position vector of magnitude 6.90 m and angle θ1 = 42.0° and a velocity vector of magnitude 6.70 m/s and angle θ2 = 25.0°. Force , of magnitude 8.90 N and angle θ3 = 28.0° acts on P. All three vectors lie in the xy plane. About the origin, what are the magnitude of (a) the angular momentum of the particle and (b) the torque acting on the particle?
To find the magnitude of the angular momentum of the particle, we need to use the formula:
Angular Momentum (L) = r x p
where r is the position vector and p is the momentum vector.
Let's first find the position vector components using the given magnitude and angle:
x-component of r = magnitude * cos(angle)
= 6.90 m * cos(42.0°)
y-component of r = magnitude * sin(angle)
= 6.90 m * sin(42.0°)
Similarly, let's find the momentum vector components:
x-component of p = magnitude * cos(angle)
= 6.70 m/s * cos(25.0°)
y-component of p = magnitude * sin(angle)
= 6.70 m/s * sin(25.0°)
Now, we can calculate the cross product of r and p to find the angular momentum:
L = (x-component of r * y-component of p) - (y-component of r * x-component of p)
Next, let's find the torque acting on the particle. The torque (τ) is given by:
Torque (τ) = r x F
where r is the position vector and F is the force vector.
Just like before, let's find the force vector components using the given magnitude and angle:
x-component of F = magnitude * cos(angle)
= 8.90 N * cos(28.0°)
y-component of F = magnitude * sin(angle)
= 8.90 N * sin(28.0°)
Now, we can calculate the cross product of r and F to find the torque:
τ = (x-component of r * y-component of F) - (y-component of r * x-component of F)
By plugging in the calculated values, you can find the magnitude of both the angular momentum (L) and the torque (τ) acting on the particle around the origin.
To calculate the magnitude of the angular momentum of the particle (a), we can use the formula:
Angular Momentum = Mass × Velocity × Radius × sin(θ)
Given data:
Particle mass (m) = 8.20 kg
Velocity magnitude (v) = 6.70 m/s
Radius magnitude (r) = 6.90 m
Angle between velocity and radius (θ2 - θ1) = 25.0° - 42.0° = -17°
First, we need to find the angle between the velocity and radius vectors in the counterclockwise direction. Since θ2 is smaller than θ1, we need to subtract them.
θ = θ2 - θ1 = 25.0° - 42.0° = -17.0°
Now, we can calculate the magnitude of the angular momentum:
Angular Momentum = (mass) × (velocity) × (radius) × sin(θ)
Angular Momentum = 8.20 kg × 6.70 m/s × 6.90 m × sin(-17.0°)
Note: Since the angle is given as -17.0°, we need to take the negative sign in the sine function.
Calculating the value:
Angular Momentum = 8.20 kg × 6.70 m/s × 6.90 m × sin(-17.0°)
Angular Momentum ≈ -153 kg·m²/s
Therefore, the magnitude of the angular momentum of the particle is approximately 153 kg·m²/s.
To calculate the torque acting on the particle (b), we can use the formula:
Torque = Force × Radius × sin(θ)
Given data:
Force magnitude (F) = 8.90 N
Radius magnitude (r) = 6.90 m
Angle between force and radius (θ3 - θ1) = 28.0° - 42.0° = -14°
Similarly, we need to find the angle between the force and radius vectors in the counterclockwise direction:
θ = θ3 - θ1 = 28.0° - 42.0° = -14.0°
Now, we can calculate the torque:
Torque = (force) × (radius) × sin(θ)
Torque = 8.90 N × 6.90 m × sin(-14.0°)
Note: Since the angle is given as -14.0°, we need to take the negative sign in the sine function.
Calculating the value:
Torque = 8.90 N × 6.90 m × sin(-14.0°)
Torque ≈ -7.77 N·m
Therefore, the magnitude of the torque acting on the particle is approximately 7.77 N·m.