A 9.0 kg cylinder rolls without slippingon a rough surface. At an instant when i

Question

A 9.0 kg cylinder rolls without slippingon a rough surface. At an instant when its center of gravity has aspeed of 11.0 m/s. (a) Determine the translational kinetic energyof its center of gravity.
1 J
(b) Determine the rotational kinetic energy about its center ofgravity.
2 J
(c) Determine its total kinetic energy
3 J (a) Determine the translational kinetic energyof its center of gravity.
1 J
(b) Determine the rotational kinetic energy about its center ofgravity.
2 J
(c) Determine its total kinetic energy
3 J

Explanation / Answer

   the translational kinetic energy of itscenter of gravity will be    KET = (1 / 2) mv2    the rotational kinetic energy about itscenter of gravity    KER = (1 / 2) I2    I = m r2    I = m r2    but as r = 0    KER = 0    total KE will be    KE = KET +KER

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.