A solid sphere of mass M, radius R rolls without slipping down an incline of angle theta. Acceleration:
Ag sin theta
Bg sin theta × (1/3)
Cg sin theta × (5/7)
DZero
Answer & Solution
Correct answer: C. g sin theta × (5/7)
For solid sphere: a = g sin theta / (1 + I/(MR²)) = g sin theta / (1 + 2/5) = g sin theta / (7/5) = (5/7) g sin theta. About 71 percent of pure sliding acceleration.
Related questions
A bicycle wheel of radius $R$ rolls without slipping at linear speed $v$. The angular speeAn ice skater spinning with arms outstretched at $\omega_1$ pulls arms in, halving her momA solid disc of mass $M$, radius $R$, rotating about its centre axis has moment of inertiaA child pushes a door of width $0.8$ m perpendicular to the door at the handle with $20$ NMoment of inertia of an annular disc (inner radius r1, outer r2) about perpendicular axis For a solid cylinder rolling without slipping, the friction force at the contact point:A merry-go-round (I = 1000 kg m²) rotates at 0.5 rad/s. A 50 kg child runs and jumps onto A solid cylinder (I = MR²/2) of mass 2 kg, radius 0.1 m, rolls without slipping at 6 m/s.