Answer
given,
L(t) = 10 - 3.5 t
mass of particle = 2 Kg
radius of the circle = 3.1 m
a) torque
τ =
τ =
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 =
ω =
ω =
A = 0.52 rad/s B = -0.182 rad/s²
(a) 10241 W
In this situation, the car is moving at constant speed: this means that its acceleration along the direction parallel to the slope is zero, and so the net force along this direction is also zero.
The equation of the forces along the parallel direction is:
where
F is the force applied to pull the car
m = 950 kg is the mass of the car
is the acceleration of gravity
is the angle of the incline
Solving for F,
Now we know that the car is moving at constant velocity of
v = 2.20 m/s
So we can find the power done by the motor during the constant speed phase as
(b) 10624 W
The maximum power is provided during the phase of acceleration, because during this phase the force applied is maximum. The acceleration of the car can be found with the equation
where
v = 2.20 m/s is the final velocity
a is the acceleration
u = 0 is the initial velocity
t = 12.0 s is the time
Solving for a,
So now the equation of the forces along the direction parallel to the incline is
And solving for F, we find the maximum force applied by the motor:
The maximum power will be applied when the velocity is maximum, v = 2.20 m/s, and so it is:
(c)
Due to the law of conservation of energy, the total energy transferred out of the motor by work must be equal to the gravitational potential energy gained by the car.
The change in potential energy of the car is:
where
m = 950 kg is the mass
is the acceleration of gravity
is the change in height, which is
where L = 1250 m is the total distance covered.
Substituting, we find the energy transferred: