ic_testtwopractice

course 201

12/14 9 pmI worked out a few problems from randomly generated tests. I was hoping you could look over my work and give me some feedback.

If an automobile of mass 771 kg is moving around a circular track 13.99 meters in diameter with angular velocity 1.11 rad/sec, then how fast is it moving and what centripetal force is acting on it?

Radius=1/2 diameter=.5(14 m)=7 m

1.11 rad/sec *7m= 7.77 m/s

Acent=v^2/r=(7.77 m/s)^2/7 m=8.6 m/s^2

Fcent=acent*mass

Fcent=8.6 m/s^2*771 kg= 6650 N

?? Is it alright to use the Acent=v^2/r formula when not in a simple harmonic motion situation?

a_cent = v^2 / r is the centripetal acceleration of an object moving on a circle of radius r at speed v.

This applies to anything moving around a circle, including a satellite in circular motion, a point on an object rotating about a fixed axis, or the reference point in the circular model of simple harmonic motion.

A uniform rod of negligible mass and length 71 cm is constrained to rotate on an axis about its center. Two masses of .51 kg are attached to the rod at a distances of 29.82 cm from the axis of rotation, one on either side of the axis. Two masses of 5.1 kg are attached at the ends of the rods An unknown uniform force is applied to the rod at a position 19.17 cm from the axis of rotation, in the plane of motion of the rod, and at an angle of 30 degrees from the rod. The force is applied as the rod rotates through .23 radians from rest, which requires 1.3 seconds. The applied force is then removed and, coasting only under the influence of friction, the rod comes to rest after rotating through 3.9 radians, which requires 19 seconds.

• Find the net torque for each of the two phases of the motion.

Tau=I*alpha

I=2(1/12*.51 kg*(.2982 m)^2)+2(1/12*5.1 kg*(.355m)^2)=.1147 kgm^2

Alpha=v^2/r

alpha is angular acceleration, which is unrelated to the centripetal acceleration v^2 / r.

alpha is the rate of change of angular velocity with respect to clock time, obtained by the same reasoning we used for linear motion.

In this case we know initial angular velocity zero, angular displacement .23 rad, `dt = 1.3 s. We can reason out average angular velocity, then final angular velocity, then change in angular velocity, which we then divide by the time interval to get angular acceleration alpha. Or we can use the equations of uniformly accelerated motion (in angular form).

omega01=0

omegaf1=.23 rad(.355 m)/1.3 sec=.063 m/s

This is the average speed elocity of a point along an arc of radius .355 m, during the 1.3 sec time interval.

omega stands for angular velocity, not the velocity of a point or object along the arc.

The average angular velocity is .23 rad / (1.3 sec) = .17 rad / sec, approx..

A point moving on a circle of radius .355 m at .17 rad / sec moves along the ard at .17 rad * .355 m / sec = .063 m / s.

omegaave1=.063 m/s-0/2=.0314 m/s

this is `dv, the change in the speed of the point along the arc

omegaAve is `dTheta / `dt.

Alpha1=(.0314 m/s)^2/.355 m=.0028 m/s^2

this is the centripetal acceleration of a point or object moving at .0314 m/s on a circle of radius .355 m

Tau1=.1147 kgm^2*.0028 m/s^2=3.21*10^-4 mN

had you reasoned out the acceleration correctly this would have given you the correct torque

omega02=.063 m/s

omegaf2=0 m/s

?? I feel like I should use the given information to calculate the velocity of the second phase, otherwise the too accelerations will be the same. I will do this instead, even though I’m not sure if it is correct.

omega02= 3.9 rad (.355 m)/19 sec=.073 m/s

omega_ave_2= 3.9 rad / 19 sec=.2 rad/s, very approximately

so omega_0_2 = .4 rad/s, approx..

omegaf2=0

omegaave2=.073 m/s-0/2=.0365 m/s

correct reasoning above would have given you omega_ave

Alpha2=(.0365 m/s)^2/.355 m=.0038 m/s^2

alpha = `dOmega / `dt, not v^2 / r

Tau2=.1147 kgm^2*.0038 m/s^2=4.36*10^-4 mN

• What is the unknown force?

Force=mass*acceleration

this applies to translational motion, not to rotational motion

Force=(2*.51 kg+2*5.1 kg)*.0028 m/s^2=.0314 N

force * moment arm = torque so force = torque / moment arm

• What is the maximum angular momentum of the system? How much of this angular momentum resides in the .51 kg mass?

Momentum=I*omega

The greatest omega found above was .073 m/s

this is a velocity v, not an angular velocity omega

Momentum=.1147 kgm^2*.073 m/s=.0084 kgm^3/s

?? this unit seems incorrect.. also, I am unsure of how to pinpoint the momentum specific to the .51 kg mass. How would I do this?

angular momentum is I * omega, not I * v. v is different for different parts of the object

What is the centripetal acceleration of a satellite orbiting at a radius of 10800 km from the center a certain planet if it is moving at 10100 m/s in that orbit? What is its orbital period (i.e., how long does it take to complete an orbit)?

Acent=v^2/r=(10100 m/s)^2/108000000 m=9.45 m/s^2

Circumference=2pi*r=2pi*10800000 m=67858401 m

67858401 m/10100 m/s=6718.7 seconds

6718.7 seconds/60seconds=112 mins, 112 mins/60 mins=1.87 hours to complete an orbit

good

A neutron star might have about five times the mass of our Sun, around 10^31 kg, packed into a very nearly perfect sphere of radius roughly 10 km. If you suddenly appeared at the surface of a Neutron star you would almost instantly become a part of that nearly perfect sphere (though you would probably be integrated nearer to your point of contact, you might think of yourself as almost instantly forming a thin coat, like a coat of paint, over the surface of the star).

Suppose you decided to just orbit the star (also fatal, but pretend you somehow managed to maintain physiological and psychological integrity).

• If you orbited at a distance of 10.164 km from the center of the star, at what rate would you orbit?

V=sqrt(GM/r)=sqrt(6.67*10^-11 Nm^2/kg^2*10^31 kg/10164 m)=2.56*10^8 m/s

This is the speed at which you would orbit (this speed is actually pretty close to the speed of light, so relativistic effects would change the result, but we ignore relativistic effects for now.