A uniform cylindrical bar mass 400 kg and radius 0.16 m rollswithout slipping al

Question

A uniform cylindrical bar mass 400 kg and radius 0.16 m rollswithout slipping along a horizontal surface at 6 m/s. Find (a) KEtranslational, (b) KE rotational, (c) KE total. (d) how high up anincline will the cylinder roll? (e) How high up a frictionlessincline would the cylinder slide? ( The angular velocity wouldremain constant, and therefore be irrelevant-the answer is the sameas for a sliding block.) (f) Compare the final total energy inparts (d) and (e). I got this question wrong on my homework. I don't understandit at all and i know something like this is going to be on my examand i really want to understand it. Can someone please help me doit. Please and thank you. A uniform cylindrical bar mass 400 kg and radius 0.16 m rollswithout slipping along a horizontal surface at 6 m/s. Find (a) KEtranslational, (b) KE rotational, (c) KE total. (d) how high up anincline will the cylinder roll? (e) How high up a frictionlessincline would the cylinder slide? ( The angular velocity wouldremain constant, and therefore be irrelevant-the answer is the sameas for a sliding block.) (f) Compare the final total energy inparts (d) and (e). I got this question wrong on my homework. I don't understandit at all and i know something like this is going to be on my examand i really want to understand it. Can someone please help me doit. Please and thank you.

Explanation / Answer

mass m = 400 kg radius r = 0.16 m translational speed v = 6m/ s moment of inertia I = ( 1/ 2) m r ^ 2                              = 5.12 kg m ^ 2 (a). KE translational = ( 1/ 2) m v ^ 2                               = 7200 J (b). KE rotational = ( 1/ 2) mv ^ 2 * ( k / r ) ^ 2 for cylinder ( k / r ) ^ 2 = 1/ 2 KE rotational = ( 1/ 4 ) m v ^ 2 = 3600 J (c). KE total = KE translational + KE rotational = 10800J (d). when cylinder rolls , total kinetic energy atbottom = potential energy at top                                                                        10800 = mgh from this height h = 2.755 m (e). KE translational at bottom = potential energy attop                                   7200 = mg h '                                          h ' = 1.836 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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