- #1
SheldonG
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I am reviewing my calculus (it has been many years). Hope this is ok to ask, amongst all the young'ins...
I was doing fine until:
1) A mass M is drawn up a straight incline of given height h by a mass m which is attached to the first mass by a string passing from it over a pulley at the top of the incline and which hangs vertically. Find the angle of the incline in order that the time of ascent be a minimum.
2) A swinging pendulum is 4 feet long is rotating at the rate of 18 deg/sec when it makes an angle of 30 with the vertical. How fast is the end of the pendulum rising or falling at that moment.
F=ma, trig functions
For 1, I calculated the net force on M as Ma = 32m - 32 M sin A. Reasoning that the greatest acceleration would also make the least time, I do:
a = 32m/M - 32 sin A.
I just treat this as a derivative (which it is), and set it to zero, solving for sin A:
sin A = m/M
However, the book (Morris Kline's calculus) gives sin A = m/2M.
-=-=-=-=-
For 2, the pendulum is sweeping out a circle, so s = rA, where a is the angle with the vertical. Differentiating, I get ds/dt = r dA/dt. From the problem data, r = 4 ft, dA/dt = 18 deg/sec = pi /10 rad/sec. ds/dt is v, the velocity, so the answer should be 4 (pi/10) = 2*pi/5 ft/sec. But the answer in the book is pi/5 ft/sec.
I thought perhaps they might want the vertical speed. For that I get vy = v/sin 30 = 2v = 4*pi/5 ft/sec. Even farther away.
Any help you can give to help this old guy keep his brain alive would be appreciated.
Thank you,
Sheldon
I was doing fine until:
Homework Statement
1) A mass M is drawn up a straight incline of given height h by a mass m which is attached to the first mass by a string passing from it over a pulley at the top of the incline and which hangs vertically. Find the angle of the incline in order that the time of ascent be a minimum.
2) A swinging pendulum is 4 feet long is rotating at the rate of 18 deg/sec when it makes an angle of 30 with the vertical. How fast is the end of the pendulum rising or falling at that moment.
Homework Equations
F=ma, trig functions
The Attempt at a Solution
For 1, I calculated the net force on M as Ma = 32m - 32 M sin A. Reasoning that the greatest acceleration would also make the least time, I do:
a = 32m/M - 32 sin A.
I just treat this as a derivative (which it is), and set it to zero, solving for sin A:
sin A = m/M
However, the book (Morris Kline's calculus) gives sin A = m/2M.
-=-=-=-=-
For 2, the pendulum is sweeping out a circle, so s = rA, where a is the angle with the vertical. Differentiating, I get ds/dt = r dA/dt. From the problem data, r = 4 ft, dA/dt = 18 deg/sec = pi /10 rad/sec. ds/dt is v, the velocity, so the answer should be 4 (pi/10) = 2*pi/5 ft/sec. But the answer in the book is pi/5 ft/sec.
I thought perhaps they might want the vertical speed. For that I get vy = v/sin 30 = 2v = 4*pi/5 ft/sec. Even farther away.
Any help you can give to help this old guy keep his brain alive would be appreciated.
Thank you,
Sheldon