Question

A 2.0 kg particle moves in a circle of radius 3.1 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 - (3.5 N·m)t.
(a) Find the magnitude and direction of the torque acting on the particle.
? N·m
Is it upward or downward?
(b) Find the angular velocity of the particle as a function of time in the form ?(t) = A + Bt.
A = ?
B = ?

Answer 1

Answer:

Explanation:

mass, m = 2 kg

radius, r = 3.1 m

L(t) = 10 - 3.5 t

(a) $\tau =\frac{dL}{dt}$

So, differentiate angular momentum with respect to time

τ = - 3.5 Nm

As the particle is rotating clockwise, so the angular velocity is downward and the angular momentum decreases so torque is upward.

(b) moment of inertia, I = m R²

I = 2 x 3.1 x 3.1 = 19.22 kgm^2

L = I x ω

where, ω is the angular velocity

ω = (10 - 3.5 t) / 19.22

ω = 0.520 - 0.182 t

So, by comparison A = 0.520 rad/s

B = - 0.182 rad/s²

Answer 2

Answer

given,

L(t) = 10 - 3.5 t

mass of particle = 2 Kg

radius of the circle = 3.1 m

a) torque

    τ = $\dfrac{dL}{dt}$

    τ = $\dfrac{d}{dt}(10 - 3.5 t)$

    τ = -3.5 N.m

Particle rotates clockwise as i look down the plane. Hence, its angular velocity is downward.

L decreases the angular acceleration upward. so, net torque is upward.

b) Moment of inertia of the particle

    I = m R^2

    I = 2 x 3.1²

    I = 19.22 kg.m²

    L = I ω

    ω = $\dfrac{L}{I}$

    ω = $\dfrac{10 - 3.5 t}{19.22}$

    ω = $0.520 - 0.182 t$

  A = 0.52 rad/s             B = -0.182 rad/s²