A 1.7 kg particle moves in a circle of radius 3.3 m. As you look down on the pla

Question

A 1.7 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 particle's angular momentum relative to the center of the circle varies with time according to L(t) = 10 N · m·s - (3.0 N · m)t.

(a) Find the magnitude and direction of the torque acting on the particle.
N · m
---Select--- upward downward

(b) Find the angular velocity of the particle as a function of time in the form (t) = A + Bt.
A =  rad/s
B =  rad/s2

Explanation / Answer

(a) We know that
Torque (T) = dL/dt
where L is angular momentum
L(t) = 10 -3t
dL/dt = -3
hence torque (T) = -3 Nm
(b) We know that
L = Iw
where I is moment of inertia and w is the angular velocity
I = mr2 where m is the mass of the particle and r is the radius of the circle
I = (1.7)*(3.3)2 = 18.513 kg-m2
w = L / I = (10 -3t) /18.513
w = 0.540 - 0.162 t
Hence A = 0.540
B = -0.162

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