Question

A machine part rotates at an angular speed of 0.07 rad/s; its speed is then increased to 3.4 rad/s at an angular acceleration of 0.50 rad/s2. (a) Find the angle through which the part rotates before reaching this final speed. 11.5 Correct: Your answer is correct. rad (b) In general, if both the initial and final angular speed are doubled at the same angular acceleration, by what factor is the angular displacement changed? Why?

Answer 1

Answer:

In first case angular displacement will be 11.55 radian

In second case it will be 4 times the previous case          

Explanation:

We have given that machine rotates at an angular velocity of 0.07 rad/sec

So initial angular velocity $\omega _i=0.07rad/sec$

And its speed is increases to 3.4 rad/sec

So final angular velocity is $\omega _f=3.4rad/sec$

Angular acceleration $\alpha =0.5rad/sec^2$

According to third equation of motion we know that

$\omega _f^2=\omega _i^2+2\alpha \Theta$

$3.4^2=0.07^2+2\times 0.5\times  \Theta$

$\Theta =11.55radian$

Now initial angular velocity and final angular velocity both are doubled but angular acceleration is constant

So $\omega _i=2\times 0.07=0.14rad/sec$

And  $\omega _f=2\times 3.4=6.8rad/sec$

So $6.8^2=0.14^2+2\times 0.5\times \Theta$

$\Theta =46.22radian$

So angular displacement will be 4 times the earlier displacement  

Answer 2

The angle through which the part rotates and what factor is the angular displacement changed is mathematically given as

∅=46.22rads

Angular displacement will be 4x the initial displacement.

What is  the angle through which the part rotates and what factor is the angular displacement changed?

Question Parameter(s):

A machine part rotates at an angular speed of 0.07 rad/s;

its speed is then increased to 3.4 rad/s

an angular acceleration of 0.50 rad/s2.

Generally, the equation for the Newtons motion law  is mathematically given as

$\omega _f^2=\omega _i^2+2\alpha \Theta$

Therefore

$3.4^2=0.07^2+2* 0.5* \theta$

∅=11.55rads

Hence

wi=2*0.07=

wi=0.14rad/sec

w_f=2*3.4

wf=6.8rad/sec

In conclusion, The angle is

6.8^2=0.14^2+2* 0.5* ∅

∅'=46.22rads

Since,

∅*4=∅'

The angular displacement will be 4x the initial displacement

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