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I am Lyosha [343]
3 years ago
14

Part A A uniform solid disk of radius 1.60 m and mass 2.30 kg rolls without slipping to the bottom of an inclined plane. If the

angular velocity of the disk is 4.46 rad/s at the bottom, what is the height of the inclined plane?
Physics
1 answer:
kifflom [539]3 years ago
3 0

Answer:

Height will be 3.8971 m        

Explanation:

We have given that radius of the solid r = 1.60 m

Mass of the solid disk m = 2.30 kg

Angular velocity \omega =4.46rad/sec

Moment of inertia is given by I=\frac{1}{2}mr^2

Transnational Kinetic energy is given by KE=\frac{1}{2}mv^2 as we know that v = v=\omega r

So KE=\frac{1}{2}m(\omega r)^2

Rotational kinetic energy is given by KE_{ROTATIONAL}=\frac{1}{2}I\omega ^2=\frac{1}{2}\left ( \frac{1}{2}mr^2 \right )\omega ^2=\frac{1}{4}m(r\omega )^2

Potential energy is given by mgh

According to energy conservation

mgh=\frac{1}{2}m(\omega r)^2+\frac{1}{4}m(\omega r)^2

h=\frac{3r^2\omega ^2}{4g}=\frac{3\times 1.60^2\times 4.46^2}{4\times 9.8}=3.8971m

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