OH NO!! I don’t know the mass???

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Presentation transcript:

OH NO!! I don’t know the mass??? PEi + KEi + Wc = PEf + KEf + Wn a) BEFORE AFTER KEf = ½mv2 PEi = KEf PEf = 0 mgh = ½mv2 OH NO!! I don’t know the mass??? KEi = 0 4.0 m PEi = mgh Wc = 0 Wn = 0 Substitute: (9.8)(4) = ½v2 v = 8.85 m/s

How do we account for that? This is a simplified scenario aside from friction, what else did we neglect? The ball is rolling How do we account for that? Rotational KE - KE = ½Iω2

Moment of Inertia Mass analog is moment of inertia, I r defined relative to rotation axis SI units are kg m2

More about Moment of Inertia I depends on both the mass and its distribution. If mass is distributed further from axis of rotation, moment of inertia will be larger.

Moment of Inertia of a Uniform Ring Divide ring into segments The radius of each segment is R

Other Moments of Inertia bicycle rim rolling bar baton swinging bar basketball bowling ball

Rotational Kinetic Energy Each point of a rigid body rotates with angular velocity w. Including the linear motion KE due to rotation KE of center-of-mass motion

Substitute in the I expressions PEi + KEi + Wc = PEf + KEf + Wn b) BEFORE AFTER KEf = ½mv2 + ½Iω2 PEf = 0 PEi = KEf mgh = ½mv2 + ½Iω2 Now what??? KEi = 0 4.0 m PEi = mgh Wc = 0 Can’t cancel the mass (yet !) Wn = 0 Substitute in the I expressions ball mgh = ½mv2 + ½( mr2)ω2 ring Now what??? v = ω.r mgh = ½mv2 + ½(mr2)ω2 so …….. ω = v disk r mgh = ½mv2 + ½(½mr2)ω2

v ω = r Algebra !!! ball mgh = ½mv2 + ½( mr2)ω2 ring disk mgh = ½mv2 + ½(½mr2)ω2 ball mgh = ½mv2 + ½( mr2)( )2 Algebra !!! ring mgh = ½mv2 + ½(mr2)( )2 disk mgh = ½mv2 + ½(½mr2)( )2

Now cancel the m’s ball mgh = ½mv2 + ½(mr2)( ) mgh = ½mv2 + ½( mv2) ring mgh = ½mv2 + ½(mr2) mgh = ½mv2 + ½(mv2) disk mgh = ½mv2 + ½(½mr2) mgh = ½mv2 + ½(½mv2) ball mgh = mv2 mgh = ½mv2 + mv2 Now cancel the m’s ring mgh = mv2 mgh = ½mv2 + ½mv2 disk mgh = ¾mv2 mgh = ½mv2 + ¼mv2

ball (9.8)(4) = v2 vball = 7.48 m/s gh = v2 ring (9.8)(4) = v2 vring = 6.26 m/s gh = ¼v2 disk (9.8)(4) = ¾v2 vdisk = 7.22 m/s gh = ¾v2