Question 7.6: Goal Apply the concepts of centripetal acceleration and tang...
Goal Apply the concepts of centripetal acceleration and tangential speed.
Problem A race car accelerates uniformly from a speed of 40.0 m/s to a speed of 60.0 m/s in 5.00 s while traveling counterclockwise around a circular track of radius 4.00 × 10² m. When the car reaches a speed of 50.0 m/s, find (a) the magnitude of the car’s centripetal acceleration (b) the angular speed (c) the tangential acceleration, and (d) the magnitude of the total acceleration.
Strategy Substitute values into the definitions of centripetal acceleration (Equation 7.13), tangential speed (Equation 7.10), and total acceleration (Equation 7.18). Dividing the change in linear speed by the time yields the tangential acceleration.
v_t = rω [7.10]
a_c = \frac{v^2}{r} [7.13]
a = \sqrt{{a_t}^2 + {a_c}^2 } [7.18]
(a) Find the magnitude of the centripetal acceleration when v = 50.0 m/s.
Substitute into Equation 7.13:
a_c = \frac{v^2}{r} = \frac{(50.0 m/s)^2}{(4.00 × 10^2 m)} = 6.25 m/s^2
(b) Find the angular speed.
Solve Equation 7.10 for ω and substitute:
ω = \frac{v}{r} = \frac{50.0 m/s}{4.00 × 10^2 m} = 0.125 rad/s
(c) Find the tangential acceleration.
Divide the change in linear speed by the time:
a_t = \frac{v_f – v_i}{Δt} = \frac{60.0 m/s – 40.0 m/s}{5.00 s} = 4.00 m/s
(d) Find the magnitude of the total acceleration.
Substitute into Equation 7.18:
a = \sqrt{{a_t}^2 + {a_c}^2 } = \sqrt{(4.00 m/s^2)^2 + (6.25 m/s^2)^2}
a = 7.42 m/s^2
Remarks We can also find the centripetal acceleration by substituting the derived value of ω into Equation 7.17.
a_t = \frac{r^2ω^2}{r} = rω^2 [7.17]