HRWP8p297

Problem 8 contains the relation, r << R, which means "r is much less than R" (see page A-9 for meanings of symbols). That means that you can approximate the smaller object as a point mass in relation to the larger object. In problem 8, this means that you can assume that the ball's center-of-mass rises 2R from the bottom of the loop to the top. In reality, the center-of-mass only rises 2R-2r because the CM is at r at the bottom of the loop and is r below the top of the loop when it is at the top.

HRWP15p298

The linear acceleration of the yo-yo is the same as the center-of-mass acceleration (because we assume the yo-yo is going straight down). You also have to assume that the string is rolling directly off the axle. That way, when you use the relationship which ties angular motion to linear motion, a = α r, you use r = radius of axle, not the radius of the yo-yo.

Using the slope of a graph to solve problems.

There are several quantities in mechanics which are the derivative of another quantity. Some are:

1) the net force on an object is the derivative of the momentum of that object with respect to (wrt) time,
2) the net force on an object is the derivative of that object's potential energy with respect to position.
3) velocity is the derivative of position wrt time
4) acceleration is the derivative of velocity wrt time

To see how to use this information, let's use #1 as an example. If you are shown a graph of momentum (y axis) v.s. time (x axis), then the force on that object is the slope of the line. For #2, it would be a graph of potential energy (y) v.s. position (x). An example of the latter is HRW problem #37 on page 192.

HRWP37p192

For part a, you need to know the initial kinetic energy so that you can compare it to the potential energy graph. If there is left over energy at 1 m (delta U - KE > 0) then that energy is kinetic and you can derive the speed for x < 2 m.
For part b, the negative of the slope of the line is the force. The sign gives you the direction (magnitude is always positive).

Remember that U + KE = constant (since no non-conservative forces are present). KE = .5 mv².

HRWP49p79

In uniform circular motion, the magnitude of the acceleration is v²/r. Since v = 2π r/T then a = 4π² r/T². Now we see that for any object on the merry-go-round, the magnitude of the acceleration depends linearly on r. Since multiplying a vector by a scalar does not change the direction of the vector, objects along a single radial line will have accelerations which only differ by a scalar quantity.

HRWP36p78

To solve this problem, you have to do a bunch of steps. I'm listing the steps to part a in reverse order since that is how you would need to tackle the problem:

1. To find d, you need the horizontal speed (v_x) since
   d = (v_x) t and t=4s.


2. To find v_x, you need the total velocity (v) when it lands on
   the roof (because v_x = v cos(60°))


3. To find v, you need to find the vertical velocity (v_y) when it
   lands on the roof.


4. To find the final vertical velocity (v_y), you need the initial
   vertical velocity (v_y0).


5. To find v_y0, use the acceleration equation:
   
   h = .5 g t² + v_y0 t


6. Now, you can find v_y since v_y = v_y0 + gt


7. Next, you can find v since v_y = v sin(60°)


8. Going backwards, you can find d.


9. Parts b and c are easy since you already have v_y0 and
   v_x (it's the usual translation from cartesian to polar coordinates).
   

HRWP33p78

1. Using the distance and time, you can find the horizontal velocity (which is assumed to be constant).
2. Using the angle of impact, you can find the final vertical velocity.
3. "Running the tape backwards", you can use the usual vertical equation along with the time to get h.
4. Using the time and the final vertical velocity, you can get the initial vertical velocity (still running the tape backwards).
5. You can now get the total initial velocity and the initial angle.

HRWP31p78

1. Assume an angle, theta.
2. Split the initial velocity into vertical and horizontal components (sine and cosine).
3. Write down the horizontal equation (using the V_x from step 2).
4. Write down the usual vertical equation but h = 0 (bullet starts and ends at same height).
5. Solve horizontal equation for t.
6. Plug t into vertical equation, use some trig identities, and solve for theta.
7. Finally, solve for distance above target (more simple trig).

HRWP28p78

The time the ball is in the air depends on the initial vertical component of the velocity and on gravity so you have to split the initial velocity into vertical and horizontal components. The vertical gives you the time in the air and the horizontal tells you how far away it lands. That gives you the soccer player's required distance and you already have the time.

HRWP22p77

You have the initial velocity. Use sine and cosine to break it into components (vx (ihat) + vy (jhat)). Now, you can ignore the horizontal part and just the vertical velocity to find the height at 5.5s (part a). You can also find the vertical (negative) velocity at that time. Combine with the constant horizontal velocity to get the total velocity at that time (part b). Finally, you can find out how long it takes to reach it's maximum height (zero velocity) and calculate how high it will be (a vertical-only equation but watch out for signs).