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maxonik [38]
3 years ago
14

What is Angular acceleration, please explain this concept. Give any equations that involve angular acceleration and explain them

.

Physics
1 answer:
Eva8 [605]3 years ago
8 0

Answer:

Angular acceleration is defined as the rate of change of angular velocity of a body.

consider the attached figure as shown

It rotates with an angular velocity \omega

An point inside the object rotates along the path as indicated thus turning by an angle \theta in time 't'

Thus we have

\alpha =\frac{d\omega }{dt}\\\\=\frac{d}{dt}(\frac{d\theta }{dt})\\\\\therefore \alpha =\frac{d^{2}\theta }{dt^{2}}

physically angular acceleration can be understood as the rate at which the angular speed of any object is changing with time.

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Suppose you first walk 12.0 m in a direction 20? west of north and then 20.0 m in a direction 40.0? south of west. how far are y
Gnesinka [82]
The representation of this problem is shown in Figure 1. So our goal is to find the vector \overrightarrow{R}. From the figure we know that:

\left | \overrightarrow{A} \right |=12m \\ \\ \left | \overrightarrow{B} \right |=20m \\ \\ \theta_{A}=20^{\circ} \\ \\ \theta_{B}=40^{\circ}

From geometry, we know that:

\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}

Then using vector decomposition into components:

For \ A: \\ \\ A_x=-\left | \overrightarrow{A} \right |sin\theta_A=-12sin(20^{\circ})=-4.10 \\ \\ A_y=\left | \overrightarrow{A} \right |cos\theta_A=12cos(20^{\circ})=11.27 \\ \\ \\ For \ B: \\ \\ B_x=-\left | \overrightarrow{B} \right |cos\theta_B=-20cos(40^{\circ})=-15.32 \\ \\ B_y=-\left | \overrightarrow{B} \right |sin\theta_B=-20sin(40^{\circ})=-12.85

Therefore:

R_x=A_x+B_x=-4.10-15.32=-19.42m \\ \\ R_y=A_y+B_y=11.27-12.85=-1.58m

So if you want to find out <span>how far are you from your starting point you need to know the magnitude of the vector \overrightarrow{R}, that is:
</span>
\left | \overrightarrow{R} \right |=&#10;\sqrt{R_x^2+R_y^2}=\sqrt{(-19.42)^2+(-1.58)^2}=\boxed{19.48m}

Finally, let's find the <span>compass direction of a line connecting your starting point to your final position. What we are looking for here is an angle that is shown in Figure 2 which is an angle defined with respect to the positive x-axis. Therefore:

</span>\theta_R=180^{\circ}+tan^{-1}(\frac{\left | R_y \right |}{\left | R_x \right |}) \\ \\ \theta_R=180^{\circ}+tan^{-1}(\frac{1.58}{19.42}) \\ \\ \theta_R=180^{\circ}+4.65^{\circ}=185.85^{\circ}


6 0
3 years ago
Ted William drops a ball from 14.5 meter to a desk that is 1.9 meters tall. What is the final speed of the ball right before it
makkiz [27]

Answer:

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 Here, we have been given the height through which the ball drops;

 Height of drop = 14.5m - 1.9m  = 12.6m

The right motion equation is;

      V²  = U² + 2gh

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U is the initial velocity  = 0

g is the acceleration due to gravity  = 9.8m/s²

h is the height

  Now insert the parameters and solve;

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Answer:

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