An object is rotating with an angular momentum 4.00 kg m2/s while being acted on by a constant torque 3.00 Nm. What is the angular speed of precession of the angular velocity of the object?
A.1.33 rad/s.
B.0.750 rad/s.
C.12.0 rad/s.
D.zero.
E.It depends on the moment of inertia of the object.
Divide the torque by the angular acceleration. T/L. 4.00 kg m^2/s over 3.00 Nm. A Nm is the same as kg m^2/s^2 so this is the way the units work out correctly to rad/s.
The angular speed of precession, denoted by ωp, can be determined using the formula:
ωp = τ / L
Where:
ωp is the angular speed of precession,
τ is the torque acting on the object,
L is the angular momentum of the object.
Given that the torque acting on the object is 3.00 Nm and the angular momentum is 4.00 kg m^2/s, we can substitute these values into the formula:
ωp = 3.00 Nm / 4.00 kg m^2/s
Simplifying this expression, we find:
ωp = 0.75 rad/s
Therefore, the correct answer is B. 0.750 rad/s.
To find the angular speed of precession, we need to use the equation for precession:
Ω = (τ / L)
where Ω is the angular speed of precession, τ is the torque acting on the object, and L is the angular momentum of the object.
Given:
τ = 3.00 Nm (constant torque)
L = 4.00 kg m^2/s (angular momentum)
Substituting these values into the equation, we get:
Ω = (3.00 Nm) / (4.00 kg m^2/s)
Ω = 0.75 rad/s
Therefore, the correct answer is B. 0.750 rad/s.