Question

A 60 cm diameter wheel accelerates from rest at a rate of 7 rad/s2. What is the tangential acceleration (in m/s2) of a point on the edge of the wheel?

Answer 1

The tangential acceleration of a point on the edge of a 60 cm diameter wheel with angular acceleration of 7 rad/s² is 2.1 m/s².

Step-by-step explanation:

The student's question involves calculating the tangential acceleration of a point on the edge of a wheel that is accelerating with a known angular acceleration. The diameter of the wheel is given as 60 cm, which can be converted to a radius of 0.30 m. The angular acceleration is given as 7 rad/s2.

To find the tangential acceleration, we can use the formula at = r⋅α. The tangential acceleration is the product of the radius (r) and the angular acceleration (α).

By substituting the given values into the formula, we get:

at = 0.30 m × 7 rad/s2 = 2.1 m/s2

Therefore, the tangential acceleration of a point on the edge of the wheel is 2.1 m/s2.

Answer 2

2.1 m/s²

Step-by-step explanation:

By definition, the angular acceleration, is equal to the rate of change of the angular velocity, ω:

α = Δω / Δt (1)

By definition of the angular velocity, we can express the linear velocity, v, as follows:

v = ω*r⇒ Δv = Δω*r (2)

Replacing Δω, from (1) in (2), we get:

Δv = α*Δt*r⇒ Δv/Δt = α*r (3)

By definition of linear acceleration, we can write the following expression;

a = α*r

For a point on the edge of the wheel, the linear acceleration is tangent to the rim, and is equal to the product of the angular acceleration times the distance to the center, which for a point on the edge of the wheel, is just the radius:

⇒ a = 7 rad/sec²*0.3m = 2.1 m/s²