A bicycle tire rolls without slipping along the ground with its center of mass moving with speed v . You may neglect any rolling resistance. What is the speed with respect to the ground of each of the following 4 points on the wheel?
a) at the top of the wheel
b) at the 3 o'clock position of the wheel
c) at the 9 o'clock position of the wheel
so i've been trying to figure this problem out for a while and i'm not sure how to go about it. there's a tangential velocity (v) at every one of these points but how do you take angular acceleration into account in this case? how do you figure out which direction the speed at each of the points takes?
2007-10-07 09:03:40 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
At any instant, you can think of the wheel as rotating about the spot of the tire in contact with the road. Call that contact point P.
a. The axle is the radius, r, away from P and going forward at speed v. Since the top of the wheel is 2*r away from the instantaneous pivot P, it is 2*r away from P. Therefore it is moving forward at 2*v.
b. The wheel's angular velocity is w. Draw a line from P to the 3 o'clock position. Call the length of the chord you just drew, Lc. The 3 o'clock position has an instantaneous velocity vector perpendicular to that line. The magnitude of this velocity, V3 is
V3 = w*Lc
How long is Lc? You should be able to get that with some trig.
c) Repeat the process used in b.
2007-10-07 11:00:50 · answer #1 · answered by sojsail 7 · 0⤊ 0⤋
I haven't dealt with things like that for about forty years but I think that you should forget about tangential velocity and try to work with angular velocity.
2007-10-07 09:15:38 · answer #2 · answered by Nikolas S 6 · 0⤊ 0⤋
a) is moving at 2v
b) same speed as c, but moving straight down
c) same speed as b, but moving straight up
d) (I'm assuming is the 6 o'clock), is not moving.
The speed of b and c have some trigonometric relation to v.
2007-10-07 09:57:38 · answer #3 · answered by ninjawilbur 1 · 0⤊ 0⤋