Question #49084

3. A solid disk of radius 0.50 m and a mass of 0.10 kg is moving with a linear or translational velocity of 5.00 m/s. Determine the disk’s:
a. Translational kinetic energy.
b. Rotational kinetic energy if rotating while translating.
1

Expert's answer

2014-11-18T12:54:05-0500

Answer on Question #49084, Physics, Mechanics | Kinematics | Dynamics

Question:

3. A solid disk of radius 0.50 m and a mass of 0.10 kg is moving with a linear or translational velocity of 5.00 m/s. Determine the disk’s:

a. Translational kinetic energy.

b. Rotational kinetic energy if rotating while translating.

Solution:

a) From the formula for the translational kinetic energy we have:


KEt=12mv2=0.50.1kg(5.0ms)2=1.25J.KE_t = \frac{1}{2}mv^2 = 0.5 \cdot 0.1kg \cdot \left(5.0\frac{m}{s}\right)^2 = 1.25J.


b) By the definition of the rotational kinetic energy we have:


KErot=12Iω2,KE_{rot} = \frac{1}{2}I\omega^2,


where II is the moment of inertia and ω\omega is the angular velocity.

For the solid disk the moment of inertia is


I=12mr2=0.50.1kg(0.5m)2=0.0125kgm2.I = \frac{1}{2}mr^2 = 0.5 \cdot 0.1kg \cdot (0.5m)^2 = 0.0125kg \cdot m^2.


From the relationship between angular and linear variables we know that v=rωv = r\omega. So, from this formula we can find angular velocity:


ω=vr=5.0m/s0.5m=10rads.\omega = \frac{v}{r} = \frac{5.0m/s}{0.5m} = 10\frac{rad}{s}.


Then, substituting the moment of inertia and the angular velocity to the formula for the rotational kinetic energy we obtain:


KErot=0.50.0125kgm2(10rads)2=0.625JKE_{rot} = 0.5 \cdot 0.0125kg \cdot m^2 \cdot (10\frac{rad}{s})^2 = 0.625J


Answer:

a) KEt=1.25JKE_{t} = 1.25J

b) KErot=0.625JKE_{rot} = 0.625J

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