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Svetach [21]
3 years ago
5

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 movin

g 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 = ?
Physics
2 answers:
Evgen [1.6K]3 years ago
7 0

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

Ghella [55]3 years ago
5 0

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

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