Rotational Inertia Hill Problem

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To determine the linear speed of a bicycle wheel rolling down an 8 m hill, the conservation of energy principle is applied, equating potential energy (mgh) to kinetic energy (KE). The initial kinetic energy includes both linear and rotational components, expressed as KE = linear KE + rotational KE. The moment of inertia for the wheel is given by I = 1/2mr^2, which is essential for calculating the rotational kinetic energy. The missing variable in the potential energy equation is the rotational speed of the wheel, which can be derived from the linear speed. Ultimately, the problem requires integrating both forms of kinetic energy to find the final linear speed at the bottom of the hill.
cnfsdstudent
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A bicycle wheel is rolled down a hill that is 8 m high, Determine the linear speed of the wheel at the bottom of the hill if the wheel is moving initially at .5 m/s


I= 1/2mr^2

mgh=1/2mv^2



I know it is set up like a KE equation, and I also know I am missing a variable to solve for on the mgh side. Does anyone know what that variable might be? I know it isn't torque.
 
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Hi cnfsdstudent! Welcome to PF! :smile:

(try using the X2 tag just above the Reply box :wink:)

You need KE + PE = constant, and also KE = linear KE + rotational KE. :smile:
 
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