Rotation of Rigid bodies

Question

A bicycle wheel with radius 0.3 m rotates from rest to 3 rev/s in 5 s. What is the magnitude and direction of the total acceleration vector at the edge of the wheel at 1.0 s?

magnitude:

direction:

Round your answer to the nearest 2 decimal places

Do not forget the units.

Answer 1

The statement of the problem leaves many aspects unspecified, so I will be making

a couple assumptions.

If my assumptions are wrong, hopefully you will still be able to use the general approach

to solve the problem with correct assumptions.

The total acceleration has two components:

centripetal acceleration denoted c, whose direction is always towards the center, and

tangential acceleration denoted a, whose direction is always tangential to the wheel.

The total acceleration is the vector sum of the two.

Let

r = 0.3 m be the radius of the wheel,

ω(t) be angular velocity at time t,

α(t) be angular acceleration at time t,

v(t) be tangential velocity at time t,

a(t) be tangential acceleration at time t,

c(t) be centripetal acceleration at time t.

The quantities v, a, c are actually vectors, but we will first just calculate their magnitude.

Here are some identities about circular motion, which I will not derive, but will assume you know

v(t) = rω(t)

a(t) = rα(t)

c(t) = v²(t)/r

ASSUMPTION 1: α(t) is constant

The constant α causes ω to increase from ω(0) = 0 to ω(5) = 3 rev/s, therefore

α = 3/5 rev/s² = 3/5 2π/s² = 6π/5 s^-2

And

ω(t) = αt

Under the Assumption 1 the above three identities become

v(t) = rαt

a(t) = rα

c(t) = (rαt)²/r = rα²t²

Let's calculate the actual numerical values for tangential acceleration a and the

centripetal acceleration c at the given time t = 1s

a(1) = 0.3×6π/5 = 1.13 m/s²

c(1) = 0.3×6π/5²×1² = 4.26 m/s²

Since tangential acceleration and cenripetal acelerations are perpendicular to each other,

their vector sum will have the magnitude

total acceleration = √(c² + a²) = √(1.13² + 4.26²) = 4.41 m/s²

The direction of the total acceleration depends on the direction of rotation

and the point where the total acceleration is to be measured.

ASSUMPTION 2: Direction of rotation is clockwise.

Without loss of generality assume that the wheel is rotating in the x-y plane

with the origin being its center.

The direction is most easily described at the edge point (-0.3, 0).

There the centripetal acceleration points in the positive x-direction, and

the tangential acceleration points in the positive y-direction.

So there the total acceleration vector is (4.26, 1.13).

Now suppose that the total acceleration is to be measured at some edge point, which

is at angle φ away from the negative x-axis.

Then the x-coordinate of the total acceleration becomes the sum of the x-components

of the centripetal and tangential acceleration.

And similarly for the y-coordinate.

So the total acceleration becomes the vector

(1.13sin(φ) + 4.26cos(φ), 1.13cos(φ) - 4.26sin(φ))