Question

Question 14

Question 14You rev your car's engine and watch the tachometer climb steadily from 1200 rpm to 5500 rpm in 2.7 s. (a) What is the angular acceleration of the engine? [ ] rad/s^2 (b) What is the tangential acceleration of a point on the edge of the engine's 3.5 cm diameter crankshaft? [] m/s^2 (c) How many revolutions does the engine make during this time? [] rev

Answer 1

a) Angular Accleration:

$T_i\Rightarrow \frac{1200rpm}{60sec}=20rev/sec\Rightarrow T_i=\frac{1}{20}s\ or \ 0.05s$

$T_f\Rightarrow \frac{5500rpm}{60sec}=91.6 \ rev/sec\Rightarrow T_i=\frac{1}{91.6}s\ or \ 0.011s$

$w_i=\frac{2 \pi}{T_i}=\frac{2 \pi }{0.05s}=125.66 \ rad/sec$

$w_f=\frac{2 \pi}{T_f}=\frac{2 \pi }{0.011s}=575.96 \ rad/sec$

Now, we use the kinematic equation $w_f=w_i+\alpha *\Delta t$ , where $\alpha$ =angular accleration

$575.96 \ rad/sec = (125.66 \ rad/sec)+ \alpha *2.7 s$

Use algebra to solve for $\alpha$

$\alpha= 166.78 \ rad/sec^2$

b) Tangental Accleration

r= 3.5cm=0.035m

$a_t=r\alpha$    , where a_t = tangental accleration

c) Number of revolutions during this time

Here we can use the kinematic equation $\theta _f = \theta _i +w_i*\Delta t \ + \ \frac{1}{2}\alpha (\Delta t)^2$

Plugging in all the numbers we just calculated we get,

$\theta _f =0 \ + \ (125.66 \ rad/sec)(2.7 s)+\frac{1}{2}(166.78 \ rad/sec^2)(2.7s)^2$

$\theta _f =947 \ rad\rightarrow \frac{947 \ rad}{2 \pi}=150.7 \ revolutions$

Answer 2

a) initial wi = 1200 rpm = 1200x 2pi rad / 60 sec = 125.66 rad/s
final wf = 5500 x 2pi rad / 60 sec = 575.96 rad/s

using wf - wi = alpha x t
575.96 - 125.66 = alpha x 2.7
alpha = 166.77 rad/s2

b) tangential acc. = alpha x radius
        = 166.77rad/s x (3.5 / 200)m = 2.92 m/s2

c) using wf^2 - wi^2   = 2 *alpha *theta
theta = 575.96^2   - 125.66^2 / 2 *166.77
     = 947.23 rad  
revolution = 947.23 / 2pi rev   = 150.76 rev