Question

a wheel on moving car slows uniformly from 70 rad/sec to 42 rad/sec in 4.2 sec. What is the angular acceleration of the wheel? At what angle does the wheel turn through in 4.2 seconds? How far did the car go if the radius of the wheel is 0.32 meters?

Answer 1

Answer:listen dω/dT is angular acceleration.

so 42-70/4.2= -32.2

the angular acceleration is -32.2 rad/sec^-2

Explanation:

also ∅=ωt+1/2αt^2

so  angle will be=70×4.2 + 1/2(-32.2)(4.2)^2=10

so the angle is 10 rad

Answer 2

The correct answer is the angular acceleration is -5.71 $\frac{rad}{sec^{2} }$ and the angle covered is 240.04 rad and the wheel goes 779.66 m when radius of wheel is 0.32 m.

Given: Initial angular velocity = 70 rad/sec.

Final angular velocity = 42 rad/sec.

Time taken = 4.2 sec.

Radius of wheel = 0.32 m.

To Find: The angular acceleration and at what angle does the wheel turn through in 4.2 seconds.

Solution:

$w_{f} =w_{i} +\alpha t$

42 = 70 + α(4.2)

-24 = 4.2α

α = $\frac{-24}{4.2}$

α = -5.71 $\frac{rad}{sec^{2} }$

Ф = $w_{i} t+\frac{1}{2}\alpha t^{2}$

Ф = $70(4.2)+\frac{1}{2}(-5.71)(4.2) ^{2}$

Ф = 294 - 53.95

Ф = 240.04 rad

As we know, v = rω and a = rα.

So, u = $rw_{i}$ = 0.32(70) = 224 m/sec.

a = rα = 0.32(-5.71) = -18.27 $\frac{m}{sec^{2} }$.

Applying the second equation of motion.

$s=ut+\frac{1}{2}at^{2}$

s = $224(4.2)+\frac{1}{2}*(-18.27)*(4.2) ^{2}$

$s=940.8-161-14$

s = 779.66 m

Hence, the angular acceleration is -5.71 $\frac{rad}{sec^{2} }$, the angle covered is 240.04 rad and the wheel goes 779.66 m when radius of wheel is 0.32 m.

#SPJ3