class 051103

Homework:

Introductory Problem Set 6, problems 1-2

Introductory Problem Set 7, Problems 1-6.

`T The area beneath a velocity vs. clock time trapezoid is (vf + v0) / 2 * `dt.
`T The first equation of uniformly accelerated motion on an interval is `ds = (vf + v0) / 2 * `dt.
`T The third equation of uniformly accelerated motion where velocity and position are regarded as functions of clock time is x = x0 + v0 t + 1/2 a t^2.
`F If the acceleration of an object is in the direction opposite its velocity then the object speeds up.
`T For an ideal projectile whose initial vertical velocity is zero, if we know any of the three quantities vf_y, `ds_y, `dt we can use the equations of motion to find the values of the other two.
`F If we choose up an incline as positive then an object coasting down the incline has a positive initial velocity.
`T If an object is moving in the positive direction and slowing down then its acceleration is in the negative direction.
`T If an object slides or rolls from rest down an incline under the influence of only friction, gravity, and the bending or compressive response of the incline to the gravitational force, then the direction of the net force acting on the object is down the incline.
`F If an object moves with changing speed along a uniform incline then its acceleration perpendicular to the incline is nonzero.
`T If an object rest on or coasts along an incline, the weight of the object tends to bend or compress the incline. The force exerted as the incline tends to decompress or bend back to its original shape is perpendicular to the surface of the incline.
 

`F The area beneath a velocity vs. clock time trapezoid is (vf - v0) / `dt * 2.
`F The first equation of uniformly accelerated motion on an interval is `ds = (vf - v0) / 2 * `dt.
`F The third equation of uniformly accelerated motion where velocity and position are regarded as functions of clock time is x = x0 + v0 t + 1/2 a `ds.
`T If the acceleration of an object is in the direction direction opposite its velocity then the object slows down.
`F For an ideal projectile whose initial vertical velocity is zero, if we know any of the three quantities vf_x, `ds_x, `dt we can use the equations of motion to find the values of the other two.  should be true
`T If we choose down an incline as positive then an object coasting down the incline has a positive initial velocity.
`T If an object is moving in the negative direction and slowing down then its acceleration is in the positive direction.
`F If an object slides or rolls from rest down an incline under the influence of only friction, gravity, and the bending or compressive response of the incline to the gravitational force, then the direction of the net force acting on the object is perpendicular to the incline.
`T If an object moves with changing speed along a uniform incline then its acceleration parallel to the incline is nonzero.
`F If an object rest on or coasts along an incline, the weight of the object tends to bend or compress the incline. The force exerted as the incline tends to decompress or bend back to its original shape is parallel to the surface of the incline.
 

1.  The gravitational acceleration at distance r from the center of the Earth is 9.8 m/s^2 * (R_earth / r )^2, where R_earth = 6.4 * 10^6 meters, approx.

• What is the acceleration of gravity for a space shuttle orbiting 7 * 10^5 meters above the Earth's surface?
• At this altitude the distance from the center of the Earth is 6.4 * 10^6 m + 7 * 10^5 m = 7.1 * 10^6 m.  At this distance the accel. of gravity is 9.8 m/s^2 * (6.4 * 10^6 m / (7.1 * 10^6 m) )^2 = 7.8 m/s^2 (?)
• What is the acceleration of gravity for a space shuttle orbiting 5 * 10^5 meters above the Earth's surface?
• How much work would have to be done against gravity to raise a 100 kg mass from 5 * 10^5 meters above the Earth's surface to 7 * 10^5 meters?
• Here the distance from center is 6.9 * 10^6 m and the acceleration of gravity is about 8.3 m/s^2 (?)
• By how much would the gravitational PE of the 100 kg mass change as the shuttle drops from 7 * 10^5 meters to 5 * 10^5 meters above the Earth's surface?
• The F vs. r graph will be a fairly straight segment of a larger curve, so F_ave will be close to the average of the two forces.  This average is 805 N.
• To lower the object 2 * 10^5 m against an average grav force of 805 N requires 805 N * (-2 * 10^5 m) = -1.6 * 10^8 Joules.
• If that mass was moving at 8000 m/s at the 7 * 10^5 meter altitude, and if no nonconservative forces acted on it, what would be its velocity at the 5 * 10^5 meter altitude?
• `dW_noncons = 0, whether ON or BY. So `dPE + `dKE = 0.  Thus `dKE = -`dPE = +1.6 * 10^8 J. 
• Original KE is .5 m v^2 = .5 * 100 kg * (8000 m/s)^2 = 3.2 * 10^9 J.
• So new KE is 3.2 * 10^9 J + 1.6 * 10^8 J = 3.36 * 10^9 J.
• Thus new velocity is sqrt( 2 * KE / m) = sqrt( 2 * 3.36 * 10^9 J / (100 kg) ) = 8080 m/s (check this).
• If a 100 kg mass moving at 8000 m/s at altitude 7 * 10^5 meters  receives an impulse of 30,000 Newton seconds then what KE does it attain?
• Impulse is equal to change in momentum.  So momentum changes by 30,000 N s = 30,000 kg m / s.  This is change in Mv.  So change in v is change in M v / m = 30,000 kg m/s / (100 kg) = 3000 m/s.
• So the new velocity is 11,000 m/s.
• New KE is .5 m v^2 = ... = 6.50 * 10^9 J = 6,250,000,000 Joules.
• In the absence of conservative forces, at what altitude would the 100 kg mass again be moving at 8000 m/s?
• To get back down to 8000 m/s we would have to lose enough KE to reduce KE from 6.25 * 10^9 J to 3.2 * 10^9 J, a reduction of 3.05 * 10^9 J.
• So we would have to gain 3.05 * 10^9 J of PE.
• To figure out the altitude required is complicated and we can't really do it yet.  The catch is that the force drops as we get higher, so it's tough to figure out PE change without a formula or something (something like calculus).

The ramp-and-screw system allows us to vary the height of the incline at the position of the screw by 1/32 of an inch per turn of the screw.  A steel ball was allowed to roll off a horizontal ramp from a position where the center of the ball was just beyond the edge of the ramp, and was observed to have a horizontal range of about 7 cm.

According to the lab notebook sheet handed out today, if we assume that v0 is always horizontal, what is v0 in each case?

How valid is it to assume that v0 is horizontal when the ball rolls off the edge of the incline at low velocity?

How valid is it to assume that v0 is in the same direction as the incline when the ball rolls off the edge of the incline at low velocity?

How valid is it to assume that v0 is in the same direction as the incline when the ball rolls off the edge of the incline at high velocity?