A 1.9 kg particle moves In a circle of radius 3.3 m. As you look down on the pla
ID: 1262305 • Letter: A
Question
A 1.9 kg particle moves In a circle of radius 3.3 m. As you look down on the plane of its orbit, the particle is initially moving Clockwise. If we call the clockwise direction positive, the particles angular momentum relative to the center of the circle varies with time according to L(t) = 10 N m.s- (5.0 N . m)t. (a) Find the magnitude and direction of the torque acting on the particle. N . m (b) Find the angular velocity of the particle as a function of time in the form omega(t) A + Bt. A = rad/s B = rad/s^2Explanation / Answer
a) Always the direction of the torque is that of the angular momentum of the particle. So it is easy to know the direction of the torque: with the right-hand rule we know that anti-clockwise is UP and clockwise is DOWN, looking at the plane of the orbit EDGE-ON. The torque's direction is DOWN. The Magnitude is: tau = r x F = || r || x || F || sin(Theta), but the angle between r and F is 90, so sin(Pi/2)=1. So tau = r F. But, F = m a (mass times acceleration). Now, L = r m v. But, the derivative of L is: dL(t)/dt = r m dv/dt = rma = d/dt(10 N ms - [5.0 N m] t ) = -5.0Nm. And: rF = -5.0 Nm = tau.
- 5 N m
b) The angular momentum can be writen as: L = Iw, with I the moment of inertia (mass times r^2) and w the angular velocity. So, simply:
w = L(t)/I = (10 N m s - [5.0 N m] t ) / (1.9 [Kg] * (3.3)^2 [m^2] ) rad/s = [ 0.48 - 0.24 t ] rad/s -- Because all units are correct!!
A=0.48
B=0.24
That's it!..
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