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SpyIntel [72]
4 years ago
11

What is the moment of inertia of an object that rolls without slipping down a 2.00-m-high incline starting from rest, and has a

final velocity of 6.00 m/s?
Physics
1 answer:
Paha777 [63]4 years ago
7 0
E = mgh +  \frac{1}2} m v^{2} + \frac{1}{2} I \omega^{2} = mgh +  \frac{1}2} m  r ^{2}   \omega ^{2}  + \frac{1}{2} I \omega^{2}

for a solid cylinder:  I =  \frac{1}{2} m r^{2}
for a hollow cylinder: I = mr^{2}

I will look at the case of a hollow cylinder:

E = mgh + I \omega ^{2} = constant \\ \\ I =  \frac{mgh}{  \omega^{2} }

That is as far as i get.


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Here m stands for combined mass (block + bullet),

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From the energy equation above, solve for v,

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Since m = 1 kg,

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p = 2.598 kg m/s.  

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u = 2.598 kg m/s / 8 × 10⁻³ kg

= 324.76 m/s.

Hence, the magnitude of the velocity of the block just after impact is 2.598 m/s and the original speed of the bullect is 324.76m/s.

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