Question 7.4: Goal Apply the rotational kinematics equations in tandem wit...
Goal Apply the rotational kinematics equations in tandem with tangential acceleration and speed.
Problem A compact disc rotates from rest up to an angular speed of 31.4 rad/s in a time of 0.892 s. (a) What is the angular acceleration of the disc, assuming the angular acceleration is uniform? (b) Through what angle does the disc turn while coming up to speed? (c) If the radius of the disc is 4.45 cm, find the final tangential speed of a microbe riding on the rim of the disc. (d) What is the magnitude of the tangential acceleration of the microbe at the given time?
Strategy We can solve parts (a) and (b) by applying the kinematic equations for angular speed and angular displacement (Equations 7.7 and 7.8). Multiplying the radius by the angular acceleration yields the tangential acceleration at the rim, while multiplying the radius by the angular speed gives the tangential speed at that point.
| Linear Motion with a Constant (Variables: x and v) |
Rotational Motion about a Fixed Axis with α Constant (Variables: θ and ω) |
| v = v_i + αt | ω = ω_i + αt [7.7] |
| Δx = v_it + \frac{1}{2}αt^2 | Δθ = ω_it + \frac{1}{2}αt^2 [7.8] |
(a) Find the angular acceleration.
Apply the angular velocity equation ω = ω_i + αt, taking ω_i = 0 at t = 0:
α = \frac{ω}{t} = \frac{31.4 rad/s}{0.892 s} = 35.2 rad/s^2
(b) Through what angle does the disc turn?
Use Equation 7.8 for angular displacement, with t = 0.892 s and ω_i = 0:
Δθ = ω_it + \frac{1}{2}αt^2 = \frac{1}{2}(35.2 rad/s^2)(0.892 s)^2 = 14.0 rad
(c) Find the final tangential speed of a microbe at r = 4.45 cm.
Substitute into Equation 7.10:
v_t = rω = (0.0445 m)(31.4 rad/s) = 1.40 m/s
(d) Find the tangential acceleration of the microbe at r = 4.45 cm.
Substitute into Equation 7.11:
a_t = ra = (0.0445 m)(35.2 rad/s^2) = 1.57 m/s^2
Remarks Because 2π rad = 1 rev, the angular displacement in part (b) corresponds to 2.23 rev. In general, dividing the number of revolutions, because 2π ∼ 6.