Rotational Physics: Stopping a Wheel in .3 sec

[icetea07]

Homework Statement

A kid turns his bike upside down and spins the tire at 2 revolutions per second. The spoked wheel measures 26 inches in diameter and weighs 5 lbs. The kid presses a piece of metal against the rubber tire to bring it to a halt. About how long will it take to bring the wheel to a halt if the kid presses with a force of 10 lbs?

Homework Equations

I=1/2 MR^2 ƩT = I(angular accel)

The Attempt at a Solution

I=1/2*5*13^2 = 422.5

ƩF=ma
422.5=10a
a=42.25

ω=ω+(angular Accel)t
0=12.567+42.25t
t=.3 sec

 

[tiny-tim]

hi icetea07! welcome to pf!

I=1/2 MR^2

nooo …

 

[PhanthomJay]

Homework Statement

A kid turns his bike upside down and spins the tire at 2 revolutions per second. The spoked wheel measures 26 inches in diameter and weighs 5 lbs. The kid presses a piece of metal against the rubber tire to bring it to a halt. About how long will it take to bring the wheel to a halt if the kid presses with a force of 10 lbs?

Homework Equations

I=1/2 MR^2 ƩT = I(angular accel)

The Attempt at a Solution

I=1/2*5*13^2 = 422.5

First, that is not the correct formula for the mass moment of inertia of a bike wheel, where most of its mass is concemntrated at the rim. You must look that up. Now secondly, I see you may not be familiar with imperial units of measure as most commonly used in the USA. Just as in SI you would have to divide the weight in Newtons by the acceleration of gravity (9.8 m/sec^2) to get its mass in kilograms, here you must here divide the weight in pounds by the acceleration of gravity (32.2 ft/sec^2) to get its mass in slugs. And the radius of the wheel should be measured in feet, not inches, for unit consistency.

ƩF=ma
422.5=10a
a=42.25

What is this? Your formula is T = Iα, where α is the angular acceleration you are trying to find, for I you can calculate it with the corrections noted, and now, what is the torque? Note that you will need to know the coefficient of friction between metal and rubber.