# Force to RPM

1. Aug 6, 2009

### tomisme

If i hit a gear/wheel with a certain force how do i calculate how the resultent rpm?
Theres a picture attached to help with my explanation.

thanks, Tom

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2. Aug 6, 2009

### Staff: Mentor

Welcome to the PF. The force you push on the pedal with generates a torque, which is transferred to the rear wheel via a gear ratio. That torque on the rear wheel generates a force at the contact patch with the ground. That force accelerates you and the bicycle forward according to the classic equation:

F = m * a
force [Newtons] is equal to the total mass [kilograms] multiplied by the acceleration [meters per second squared]

The acceleration changes your velocity (increases it), and you calculate the velocity using the kinematic equations of motion. Your speed corresponds to your RPM via the gear ratio.

Hope that helps.

EDIT -- BTW, there is also a retarding force due to air resistance, rolling resistance, etc. The sum of all forces equals the m * a term.

3. Aug 6, 2009

### tomisme

thanks for your welcome and a quick reply :). Its not a bike for a lack of a better example im hitting a gear with stick, and i need to know how fast the gear will spin at the end.

4. Aug 7, 2009

5. Aug 8, 2009

### nooma

the initial tangential speed can be calculated by f= (mv2)/r

i reckon it can be anyways :)

6. Aug 8, 2009

### Staff: Mentor

That depends on how long you apply the force! Berkeman gave you the equation for acceleration - acceleration times time is speed.

Berkeman's equation was the linear form (ie, for how fast a bike will go), but angular acceleration works the same way - check out Tony's link for how it works.

7. Aug 9, 2009

### bm0p700f

You first have to now the change in momentum of the of the hammer = mv - mu

Knowing this you them now the change in momentum of the wheel = change in angular momentum. From that you can the find the final angular velocity which you can translate to a tip velocity.

The other way is to work out the change in K.E of the hammer on hitting the gear wheel and that change is the change in K.e of the wheel. Method two is easier

Maths assuming the gear is a solid disc of constant density and the hammer stops on hitting the gear wheel:
Method 2

Change in K.E of hammer = (1/2) mv^2
change in K.E of gear wheel = (1/2)Iw^2 I is the moment of inertia of the wheel.

w is the angular velocity = 2*pi/T where T is the rotation period of the wheel. f is the frequency of rotation (number of revolutions per second)
Therefore delta K.E of wheel is = (1/2)*I*w^2 = 2*I*pi^2/T^2 = 2*I*f^2*pi^2 = I*pi^2*rpm^2/1800

So;
(1/2)*m*v^2 = I*pi^2*rpm^2/1800

rpm = sqrt(900*m*v^2)/(I*pi^2)

I hope this helps and that I have not made a mistake in my algebra.

Last edited: Aug 9, 2009