Running around and falling down - horizontal circles

I will be reminding you of this in class, but in solving these problems, you may NOT just reuse one of the equations we derive in class. On the AP, you will be expected to use basic physics principles to arrive at your answers. You have enough to memorize - do NOT memorize random equations that only apply in specific situations. Remember the physics principles and the relevant equations, and you can derive everything else.

The ONLY new equation thus far in this section has been a = v^2/R for centripetal acceleration. Everything else? DERIVE YOURSELF!

Solutions:
5. The key to this problem is that the hanging block is in equilibrium, which gets you that T = Mg. Your FBD of the disc should get you that T = mv^2/R. Some substitution and algebra gets you the formula given in the text.

6. Draw a FBD first (of course!)

sigma F_x = T = mv^2/R, so v = (TR/m)^(1/2) = 13.96 m/s.

7. Follow the steps from class, as this problem is almost identical. The max speed is 23.43 m/s. In the last step of solving for the maximum speed, the mass divides out, so this result IS independent of the mass of the car!

11. Since the turntable rotates 36 times per second, you can use this to find the speed of the coin. Write the 36 rpm as a fraction: 36 revolutions/1 minute, and then use conversion factors (1 rev = 2*pi*R meters, 60 s = 1 min) to get this to speed v = .4147 m/s.

Since the coin only slides off at this minimum speed, you know that F_s = F_s,max = mu*Fn.

Drawing a FBD, writing Newton's 2nd, and solving gives you mu = v^2/(gR) = .160

16. Your equations for each mass (with T1 as the inner cord, and positive x pointed towards the center of the circle since this is the direction of centripetal acceleration) MUST be:
mass 1: T1 - T2 = m₁v₁^2/R₁
mass 2: T2 = m₂v₂^2/R₂

You can also derive since v₁ = 2*pi*R₁/T (where this T is period, NOT tension), that v₁ = 2*pi*R₁*f. The same thing can be done for v₂.

Thus T2 = 4*m₂*R₂*(pi*f)^2.

and T1 = 4*m₁*R₁*(pi*f)^2 + 4*m₂*R₂*(pi*f)^2

40. acceleration is UP at 3.141 m/s^2.

76. Follow the steps we used for problem 3 in class. v = (g*r*tan(17.5 degrees))^1/2 = 29.15 m/s