1117

• Run Multiple Choice Generator at h:\shares\physics
• General College Physics:  Run Core Problems by giving the program 174 as your 'course'--enter this when asked for your 3-digit course number.   Do the problems for 11/03.
• University Physics:  Run Multiple Choice Generator for 201 and choose 'Calculus of Uniformly Accelerated Motion'.
• If you miss a problem be sure to click on Comment and explain what you do and do not understand about the question and the given solution.

Run 174

What is the formula for centripetal acceleration?

a_cent = v^2 / r

What is the formula for the gravitational acceleration at distance r from the center of a planet of mass M?

G M / r. 

We get this from the fact that the gravitational force on mass m is F_grav = G M m / r^2.  Dividing by m we get a_grav = F_grav / m = G M / r^2.

What is the condition for a circular orbit at distance r from the center of a planet of mass m?

r > radius of the planet, obviously

gravitational acceleration = centripetal acceleration

G M / r^2 = v^2 / r

(similar to grav force = cent force, or G M m / r^2 = m v^2 / r)

What is the formula for the velocity of a satellite in circular orbit at distance r from the center of a planet of mass M?

Solve G M / r^2 = v^2 / r for v.

Multiply both sides by r then take the square root to get

v = sqrt(G M / r).

What is the formula for the KE of a satellite a mass m in circular orbit at distance r from the center of a planet of mass M?

KE = .5 m v^2 so

KE_circ_orbit = .5 m ( sqrt(G M / r) ) ^ 2 = .5 G M m / r.

What is the KE per unit of satellite mass for a circular orbit?

Divide the KE of the orbit by m.  We get

KE of circular orbit per unit of mass = .5 G M / r.

What is the formula for the PE of a satellite in circular orbit at distance r from the center of a planet of mass M?

Gravitational PE relative to infinite separation of two masses M and m, separated by distance r, is

• PE = - G M m / r.

If we want to measure PE relative to Earth's surface, we can raise the graph (vertically translate the graph) until its horizontal-axis intercept occurs at r = R_earth.

It doesn't matter if we vertically shift the graph in this manner.  We get the same PE difference between two points from either graph.

What is the total energy of a satellite in a circular orbit?

Total energy is PE + KE.

Measuring PE relative to infinity, we get

• total energy = PE + KE =

- G M m / r + .5 G M m / r =

-.5 G M m / r.

Assuming a uniform angular acceleration of .2 rad / sec^2, what do we know about the motion of the half-meter stick on a die after its collision with the pearl pendulum?  What can we conclude about its velocity just after collision?

The end of the stick moved about 2 cm at a distance of about 25 cm from the axis of rotation.

25 cm along the arc would correspond to 1 radian.  2 cm was measured along a straight line, but for that short distance this is almost identical to the measurement we would get along the arc. 

We conclude that the angular displacement was 2/25 rad = .04 rad.

We now have final angular velocity 0, angular accel -.2 rad/s^2 and angular displacement .04 rad. 

We can use the equations of motion to get initial angular velocity, which comes out about .13 rad/s.

The pearl pendulum had length 25 cm.  How fast was it going at the instant just before it struck the half-meter stick?  Relative to the center of rotation, what therefore was its angular velocity?

From your data, pullback from the equilibrium position was about 7 cm.  The pendulum swung back to its equilibrium position.

Figure out how fast the pendulum had to be moving at collision, assuming no dissipation of energy.  (Suggestion:  Figure out how much its altitude changed and set m g `dy equal to .5 m v^2 to get v).

How did the angular acceleration of the loaded meter stick compare with that of the unloaded meter stick?