Question

Consider a bicycle wheel to be a ring of radius 30 cm and mass 1.5 kg. Neglect the mass of the axle and sprocket. If a force of 20 N is applied tangentially to a sprocket of radius 4 cm for 4 seconds, what linear speed does the wheel achieve, assuming it rolls without slipping?
a) 3 m/s
b) 24 m/s
c) 5.9 m/s
d) 7.1 m/s

Answer 1

To solve the problem it is necessary to apply the Torque equations and their respective definitions.

The Torque is defined as,

$\tau = I \alpha$

Where,

I=Inertial Moment

$\alpha =$ Angular acceleration

Also Torque with linear equation is defined as,

$\tau = F*d$

Where,

F = Force

d= distance

Our dates are given as,

R = 30 cm = 0.3m

m = 1.5 kg

F = 20 N

r = 4.0 cm = 0.04 m

t = 4.0s

Therefore matching two equation we have that,

$d*F = I\alpha$

For a wheel the moment inertia is defined as,

I= mR2, replacing we have

$d*F= \frac{mR^2a}{R}$

$d*F= mRa$

$a = \frac{rF}{ mR}$

$a = \frac{0.04*20}{1.5*0.3}$

$a=1.77 m/s^2$

Then the velocity of the wheel is

$V = a *t \\V=1.77*4 \\V=7.11 m/s$

Therefore the correct answer is D.