# Pebble in a Rolling Tire–Finding Velocity and Acceleration?

Find the coordinates of a pebble stuck in the tread of a rolling tire that is rotation counterclockwise (positive) with angular velocity w. The tire rolls w/o slipping on the ground. Outer radius is R. At time 0 the pebble is at the top inside of the tire.

A) v_ag(t)=-Rw (Velocity of the axle of the tire relative to a fixed point on the ground)

Use the figure provided to solve the following

B) Find the position vector of the pebble relative to the initial point of contact between the wheel and ground at a time t.

C) Find the velocity vector of the pebble with respect to a fixed point on the ground.

D)Now find the acceleration vector of the pebble with respect to a fixed point on the ground.

Update:

Actually I figured these out so

B) (-Rsin(wt)-Rwt)i_hat+(Rcos(wt)+R)j_hat

C) -Rw(1+cos(wt))i_hat-Rw(sin(wt))j_hat

D) Rw^2(sin(wt))i_hat-Rw^2(cos(wt))j_hat

But now I need help with how to find magnitude of D, since it should be independent of time

Actually I figured these out so

B) (-Rsin(wt)-Rwt)i_hat+(Rcos(wt)+R)j_hat

C) -Rw(1+cos(wt))i_hat-Rw(sin(wt))j_hat

D) Rw^2(sin(wt))i_hat-Rw^2(cos(wt))j_hat

But now I need help with how to find magnitude of D, since it should be independent of time

• ?
4 days ago

Nit-picky — I think B is wrong, and it should be

(Rsin(wt) – Rwt) i_hat + (R – Rcos(wt)) j_hat

based on the citation below and accounting for your negative direction.

But I also wonder if you shouldn’t have a little “r” in there instead of the capital “R” to better describe the location of the pebble.

C switches from “the initial point” to “a fixed point,” and so I think that’s just

v^ = -rωcos(ωt) i^ – rωsin(ωt) j^

should give the direction of the velocity vector at time t.

I’ve used the little “r” here — maybe that’s wrong.

D also deals with “a fixed point,” and so we just need the magnitude and the diretion of the centripetal acceleration.

magnitude a = ω²r

so since the acceleration vector is perpendicular to the velocity vector,

a^ = ω²r * [sin(ωt) i^ – cos(ωt) j^]

You can email me if you have a question or comment.

Hope this helps.

• Nick
2 years ago

I know this is an old thread – but for the final part (the magnitude) – the answer is simply Rw^2

• oldschool
4 days ago

It will be the vector sum of the perpendicular accelerations.

One is tangential acceleration = a = ΔV/Δt = R*α = R*Δω/Δt

and the other is centripetal acceleration V²/R.

Acceleration = √(α²+a²)