Question

Pulling a string to accelerate a wheel

Answer 1

a bicycle wheel is mounted on a fixed, frictionless axle. A massless string is wound around the wheel's rim, and a constant horizontal force F of magnitude F starts pulling the string from the top of the wheel starting at time t=0 when the wheel is not rotating. Suppose that at some later time t the string has been pulled through a distance d. The wheel has moment of inertia \[I_w=kmr^2\]where k is a dimensionless number less than 1, m is th wheel's mass, and r is it's radius. Assume that the string does not slip on the wheel. Find the angular acceleration alpha of the wheel, magnitude of the angular velocity omega of the wheel when the string has been pulled a distance d, and find the speed v of the string after it has been pulled by F over a distance d.

Answer 2

so i got the first part and i just need a little help for the second and third parts. i got the first by\[\sum \tau:I_w=kmr^2\]\[F*r=kmr^2*\alpha\]\[F=\alpha*kmr\]\[\alpha=\frac{F}{kmr}\]

Answer 3

for the second part i gathered that the angular velocity has to be the same as regular straight velocity since the string doesn't slip or anything so i just did some kinematics and found time\[d=v_0t+\frac{1}{2}at^2\]\[d=\frac{1}{2}at^2\]\[t=\sqrt{\frac{2d}{a}}\]so now that i have this i'm stuck here...

Answer 4

@surry99

Answer 5

Hi, let me take a look hang on please

Answer 6

at time t, the instantaneous velocity of the string must equal the tangential velocity of the disc....agreed?

Answer 7

yeah. agreed

Answer 8

ok so lets find the angular velocity omega..
by defintion :
omega = velocity tangential/r
now assuming the rope does not stretch, if it moves a distance d, do you see the that must be equal to the arc length that the disc rotated through in time t?

Answer 9

yeah. it has to be the same like the question that you helped me with before right?

Answer 10

similar indeed...
so w = v/r but v = delta s/delta t but since it started from rest t=0, when s=o
therefore omega = v/r = s/rt

Answer 11

cripes. i have a class in 8 minutes. i'll be back later tonight. sorry.

Answer 12

no problem at all...TTYL

Answer 13

I will be on tonight and we can finish this up.

Answer 14

i actually solved this problem on my own so i'm going to close this. thanks for the help though. i'll ask you for more help if i come into more hard questions. hope you don't mind. :P

Answer 15

ok great!