Question

(a) What is the tangential acceleration of a bug on the rim of a 6.0 in. diameter disk if the disk moves from rest to an angular speed of 79 revolutions per minute in 4.0 s? m/s2 (b) When the disk is at its final speed, what is the tangential velocity of the bug? m/s (c) One second after the bug starts from rest, what is its tangential acceleration? m/s2 What is its centripetal acceleration? m/s2 What is its total acceleration? m/s2 ° (relative to the tangential acceleration)

Answer 1

(a) Apply the formula –

ω_f = ω_i + α(t)
ω_f = final angular velocity = (79 rev/min)(2π rad/rev)(1min/60s) = 8.27 rad/s
ω_i = initial angular velocity = 0 rad/s
t = time = 4.0 s

The ω_i term is eliminated, so we'll just say ω_f = ω for simplicity. Solve for α:
α = ω / t
α = (8.27 rad/s) / (4.0 s)
α = 2.07 rad/s²

(b) v = rω
v = velocity
r = radius = 3.0 in

v = (3.0 in)(8.27 rad/s)
v = 24.81 in/s
v = 2.07 ft/s

(c) a_t = rα
a_t = (3.0 in)(2.07 rad/s²)
a_t = 6.21 in/s²
a_t = 0.52 ft/s²

Again, for the next part, we have to find the angular velocity at t = 1s first.
ω_f = ω_i + α(t)
ω_f = 0 + (2.07 rad/s²)(1 s)
ω_f = ω = 2.07 rad/s

a_n = rω²
a_n = (3.0 in)(2.07 rad/s)²
a_n = 12.85 in/s²
a_n = 1.08 ft/s²

For total acceleration, use the pythagorean theorem:
a = √[(a_n)² + (a_t)²]
a = 14.27 in/s²
a = 1.19 ft/s²

Finally, to find the angle:
Δθ = ω(t) + (1/2)α(t)²

Δθ = change in angle
ω = initial angular velocity = 0 rad/s

Δθ = (0 rad/s)(1 s) + (1/2)(2.07 rad/s²)(1 s)²
Δθ = 1.03 rad
Δθ = 59.0°