A bicycle wheel of radius $R$ rolls without slipping at linear speed $v$. The angular speed $\omega$ is:
A$vR$, the product of speed and radius for the wheel
B$v + R$, the sum of linear and length scales here
C$v/R$, from $v = \omega R$ rolling constraint
D$R/v$, the inverse ratio under rolling
Answer & Solution
Correct answer: C. $v/R$, from $v = \omega R$ rolling constraint
Rolling constraint: $v = \omega R$, so $\omega = v/R$.
Related questions
An 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. For a body rotating at constant angular velocity omega, the linear speed at distance r fro