2007-12-04
Today's nonsense
- We have homework due tomorrow.
- Katie and Dakota and Alyssa are alergic to pine trees. Dakota and Katie had a speech impediment. No correlation. "You're the derivative of an alergic."
- Please note that it is Kandy with a K. So if you're Ms. Dees and you call and you ask for candy, know that you're going to be asking for the candy in the back of the room all in the boxes.
ex 2) You swing a pail of water in a vertical circle of radius r. The speed of the pail is v_(t) at the top of the circle. (a) Find the force exerted on the water by the pail at the top of the circle. (b) Find the minimum value of v_(t) for the water to remain in the pail. (c) Find the force exerted by the pail on the water at the bottom of the circle, where the pail's speed is v_(b).
Tangental acceleration later?
Sum of F_(y)= -F - mg = ma_(y) = m (-1)v^(2) / r
F = ma_(y) + mg
F = m(-1)v^(2)/r - mg
F = m(-v^(2) / r - g)
a = (-v^(2) / r - g)
(b) Find the minimum value of v_(t) for the water to remain in the pail. What's the minimum velocity at which you can swing this thing so that the water stays in the bucket? Your minimum velocity is going to be when your force is as at an absolute minimum, and what's the smallest that a force can get?
minimum where F=0
(-v^(2) / r - g) = 0
-v^(2) = 9.81r
v = sqrt(9.81r)
The smaller the radius, the smaller the velocity (because of the square root).
(c) Find the force exerted by the pail on the water at the bottom of the circle, where the pail's speed is v_(b).
Sum of F_(y) = F - mg = ma_(y) = mv^(2) / r ... and that's a positive v^2 because the acceleration is upwards, towards the center of the circle.
We're looking for force. F = mg + mv^(2) / r
Put it into the terms that you already know.
ex 3) A tether ball of mass m is suspended from a rope of length L and travels at constant speed v in a horizontal circle of radius r. The rope makes an angle theta given by sin(theta) = r/L, as shown in Figure 5-28. Find (a) the tension in the rope, and (b) the speed of the ball.
The angle of the rope to the supporting beam of the rope, hanging over the circle, is theta. The "horizontal circle" is a disk on a plane. The rope has a tension. Down is mg.
Sum of F_(x) = Tsin(theta) = ma_(x) = mv^(2) / r
Should that be negative Tsin(theta)? Either direction you can call negative. We can call "to the left" as negative, or we can think of the tether ball on the other side of the pole. If we call T negative, then the a_(x) would have to be negative.
Sum of F_(y) = Tcos(theta) - mg = ma_(y) = 0
Find the tension in the rope, solve for T.
mgtan(theta) = (1/r)mv^(2)
V = sqrt(rgtan(theta)
ex 6) A pilot of mass m = 56 kg comes out of a vertical dive in a circular arc of radius 420 m. Her speed at the bottom of the circle is 200 km/h. (a) What's the magnitude of he(r) acceleration? So draw a half circle, with at the bottom point we have our free-body diagram, and there's mg going down, and F going up, and so ...
Sum of F_(y) = F - mg = ma_(y) = mv^(2) / r
a_(y) = v^(2) / r = (55.556 m/s^2) / 420 m = 7.349 m/sec^2
(b) What's the net force acting on her at the bottom of the circle?
F_(net) = ma = 7.349 * 56 kg = 411.523 N
(c) What's the force exerted by the airplane seat on her?
We're looking for the original F from the free-body diagram. Keep in mind that the sum of forces gives us the net force which equals F - mg. So we're looking for F. We have mg, we have F_(net).
F = 960.883 N = F - mg.
We have an object on a string, and we're swinging it over head but the rope is at an angle, ...
ex 7)
m = 326 g
d_(string) = 1.4 m
theta = 35 deg with the horizontal
v = ?
a = v^(2)/r and v = 2 * pi * r / T.
cos(theta) = r / d
r = d cos(theta)
r = 1.4cos(theta)
v = 2 * pi * 1.4cos(theta) / T
Sum of F_(y) = Tsin(theta) - mg = 0
Sum of F_(x) = Tcos(theta) = ma_(x) = mv^(2) / r
T = mg/sin(theta) = (1/r)(1/cos(theta))mv^(2)
v^(2) = g(1.4cos(theta))cos(theta)/sin(theta)
v = sqrt (g(1.4cos(theta))cos(theta)/sin(theta))
ex 5) A curve of radius 30 m is banked at an angle theta. Find theta for which a car can round the curve at 40 km/h even if the road is frictionless.