General Equation for Acceleration

Question

Question: A drum of radius R rolls down a slope without slipping. Its axis has acceleration a parallel to the slope. What is the drum’s angular acceleration α?

To solve this question, I used the general equation for acceleration:

a = (r¨)rˆ + (r˙)(θ˙)θˆ + (r˙)(θ˙)θˆ + r(θ¨)θˆ − r((θ˙)^2) rˆ

where r is the radius, (r˙) is the derivative of r, (r¨) is the second derivative of r, (θ˙) is the angular velocity, (θ¨) is the angular acceleration and rˆ and θˆ are unit vectors.

In this example, r is a constant meaning both the first and second derivative is 0 which simplifies the equation to:

a = r(θ¨)θˆ − r((θ˙)^2)rˆ = r(α)θˆ − r(ω^2)rˆ

My solution leads me to have an extra term.

In the solution manual, the answer is simply:

a = rα which implies α = a/r.

What am I doing wrong?

Answer 1

Let $b(t)$ be the distance that the barrel has rolled at time $t$. Then $b(t) = 1/2 at^2 + v_0 t$.

At time $t$, assuming there is no slip between the barrel and the surface, the point on the barrel that is touching the surface (call it particle $p_t$) must be moving with velocity exactly negative of the barrel's velocity $v_{p_t}(t) = -b'(t) = - a t -v_0$.

Its angular velocity is $\omega_{p_t}(t) = v_{p_t}(t)/r = (-a t - v_0)/r$.

But the angular velocity is uniform for all particles on the barrel so we can drop the $p_t$ and just write $\omega(t) = (-a t - v_0)/r$

Now just take the derivative $\omega'(t) = -a/r$ to get the angular acceleration.

Answer 2

Note that angular acceleration is: $$\alpha = \frac{d\omega}{dt} = \frac{d^2\theta}{dt^2}$$ If we define $v$ to be the velocity of the axis perpendicular to the slope, with $s$ being the distance travelled: $$v = \frac{ds}{dt}$$ In order for an object to travel a distance $s$ while having angular velocity, without slipping, we have(r is the radius of the rolling object): $$v = \omega r$$ This is easy to show, and proof can be found at: http://hyperphysics.phy-astr.gsu.edu/hbase/rotq.html , however this is not being asked here, so it won't be proven by me. Obviously, then we have: $$\frac{dv}{dt} = a = r \frac{d\omega}{dt} = r\alpha $$