A 10.0 kg cylinder rolls without slipping on a rough surface. At an instant when its center of gravity has a speed of 10.0 m/s, determine: 1. The translational kinetic energy of its center of gravity. 2. The rotational kinetic energy about its center of gravity. 3. Its total kinetic energy
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The Correct Answer and Explanation is:
To solve this problem, we need to recall the basic formulas for kinetic energy, both translational and rotational.
Given:
- Mass of the cylinder, m = 10.0 kg
- Velocity of center of gravity, v = 10.0 m/s
- The cylinder rolls without slipping, so the condition v=rωv = r\omegav=rω applies
- For a solid cylinder, the moment of inertia about its center of mass is I=12mr2I = \frac{1}{2}mr^2I=21mr2
1. Translational Kinetic Energy
The translational kinetic energy is given by:KEtrans=12mv2KE_{trans} = \frac{1}{2}mv^2KEtrans=21mv2
Substituting values:KEtrans=12(10.0)(10.0)2=12(10.0)(100)=500 JKE_{trans} = \frac{1}{2}(10.0)(10.0)^2 = \frac{1}{2}(10.0)(100) = 500 \text{ J}KEtrans=21(10.0)(10.0)2=21(10.0)(100)=500 J
2. Rotational Kinetic Energy about the Center of Gravity
Rotational kinetic energy is:KErot=12Iω2KE_{rot} = \frac{1}{2}I\omega^2KErot=21Iω2
For a solid cylinder, I=12mr2I = \frac{1}{2}mr^2I=21mr2 and using v=rω⇒ω=vrv = r\omega \Rightarrow \omega = \frac{v}{r}v=rω⇒ω=rv, we get:KErot=12⋅12mr2⋅(vr)2=14mv2KE_{rot} = \frac{1}{2} \cdot \frac{1}{2}mr^2 \cdot \left(\frac{v}{r}\right)^2 = \frac{1}{4}mv^2KErot=21⋅21mr2⋅(rv)2=41mv2
Now plug in values:KErot=14(10.0)(10.0)2=14(10.0)(100)=250 JKE_{rot} = \frac{1}{4}(10.0)(10.0)^2 = \frac{1}{4}(10.0)(100) = 250 \text{ J}KErot=41(10.0)(10.0)2=41(10.0)(100)=250 J
3. Total Kinetic Energy
KEtotal=KEtrans+KErot=500+250=750 JKE_{total} = KE_{trans} + KE_{rot} = 500 + 250 = 750 \text{ J}KEtotal=KEtrans+KErot=500+250=750 J
Final Answers:
- Translational kinetic energy = 500 J
- Rotational kinetic energy = 250 J
- Total kinetic energy = 750 J
Explanation:
When an object like a cylinder rolls without slipping, it has two types of kinetic energy: translational (from its forward motion) and rotational (from spinning). These energies are additive. The “no slipping” condition ties the rotational speed to the translational speed, letting us express rotational energy in terms of velocity. Since it is a solid cylinder, its moment of inertia is lower than that of a hollow one, meaning less energy goes into rotation and more into forward motion. The total kinetic energy reflects both effects combined.
