Question

Chuck Norris’s roundhouse kick is quite impressive. Assume his foot begins from rest and takes two thirds of a second to swing completely around a 1.1-m-radius circular arc. His foot has a constant angular acceleration for the first third of a second and then zero after that.
A. If the angular acceleration is assumed constant for the first third of a second, and then zero after that, what is its magnitude during the first third of a second?
B. What is the foot's tangential acceleration during the first 0.334 seconds?
C. What is the maximum tangential speed his foot obtains?
D. When his foot reaches maximum tangential speed as found in part c, what percent of full rotation does this occur?

Answer 1

Answer:

A) $\alpha$  = 37.68 rad/$s^{2}$

B) a = 41.45 m/$s^{2}$ = Tangential acceleration

C) v = 13.68 m/s

Maximum Tangential Speed

D) % age of full circle rotation = 32.64%

Explanation:

Solution:

A)

For the first part of the circle

Ф1 = $w_{0}$$t_{1}$ + 1/2$\alpha t1 ^{2}$            

As initial speed is zero so,

Ф1 = 0 + 1/2$\alpha (0.33) ^{2}$  

Ф1 = 0.05445$\alpha$     equation 1

And final speed will be:

w = $w_{0}$ + $\alpha$$t_{1}$

w = 0 + $\alpha$ (0.33)

w = 0.33 $\alpha$    equation 2

Similarly, for second part of the circle.

Ф2 = w$t_{2}$

(2$\pi$ - Ф1) = (0.33$\alpha$) (t-$t_{1}$)

(2$\pi$ - 0.05445$\alpha$ ) = (0.33$\alpha$ ) (0.67 - 0.33)

Solving the above equation for $\alpha$, we get:

$\alpha$  = 37.68 rad/$s^{2}$

B) For tangential acceleration, we know that:

a = r$\alpha$

r = 1.1 m

a = (1.1) (37.68)

a = 41.45 m/$s^{2}$ = Tangential acceleration

C) For maximum tangential speed, we also know that,

v = rw

r = 1.1 m

w = 0.33$\alpha$

w = 0.33 x (37.68)

So, putting in the equation and solving for maximum tangential speed v:

v = rw

v = (1.1) (37.68 x 0.33)

v = 13.68 m/s

Maximum Tangential Speed

D) For percent of full rotation, we have:

Equation 1 from above:

Ф1 = 0.05445$\alpha$

Putting the values, we get the displacement.

Ф1  = 0.05445 x (37.68)

Ф1  = 2.05 rad

And we also know that, the % age of full circle rotation = Ф1/2$\pi$ x 100

% age of full circle rotation = 2.05/(2 x 3.14) x100

% age of full circle rotation = 32.64%

Answer 2

The angular kinematics allows to find it the results on the motion questions are:

      a) The angular acceleration is:     α = 137.7 rad/s²

      b) The tangential acceleration is: a = 41.5 m/s²

      c) the maximum tangential speed is: v= 13.8 m/s

      d) The fraction of angle rotated is  θ% = 33.2%

Given parameters.

• Part of rest.

• The turning time of the arc is t= ⅔ s

• The radius is r = 1.1 m

• The acceleration is constant until t₁ = ⅓ s and then it is zero.

To find.

    a) The angular acceleration.

    b) The tangential velocity for t= 0.334 s.

    c) the maximum tangential speed.

    d) the angle when the maximum speeds.

Rotational kinematics.

The rotational kinematics establishes relationship between the angular position, the angular velocity and the angular acceleration of a body.

a) The foot starts from rest so its initial angular velocity is zero.

          θ₁= w₀ + ½ α t²

          θ₁ = ½ α t₁²

The final velocity for this interval is

          w₂ = w₀ + α t

           w₂ = α t₁

After this time the acceleration is zero, so the motion is circular uniform.

           $w = \frac{\theta_2}{t_2}$  

           $\theta_2 = w \ t_2$  

Let's write our equations.

         $\theta_1 = \frac{1}{2} \alpha \ t_1^2 \\\theta_2 = w \ (t-t_1)\\w_2 = \alpha\ t_1$

       

The angle of a circle is 2π radians.

          θ₁ + θ₂ = 2 π  

          2π= ½ α t₁² + (α t₁) (t - t₁)  

           2π = α t₁  ($\frac{1}{2} \ t_1 + (t-t_1)$ )

           $2 \pi = \alpha \ t_1 ( t -\frac{t_1}{2})$

We calculate.

           $2 \pi = \alpha \frac{1}{3} ( \frac{2}{3} - \frac{1}{2} \frac{1}{3})$

            $2 \pi = \lapha \frac{1}{6} \\\alpha = 12 \pi$

           

            α = 37.7 rad/s²

Relationship linear and rotational movement.

b) Angular and linear variables are related

 the tangential acceleration is

         a = α r

           a = 37.7 1.1

           a = 41.5 m/s²

c) After the acceleration drops to zero, the velocity remains constant.

          v =w r

          v =( α t₁) r

Let's calculate.

        v= 37.7  ⅓  1.1

        v= 13.8m/s

d) They ask the percentage of rotation when it reaches this speed.

Let's find the angle rotated for the time t₁.

         θ₁ = ½ α t₁²

         θ₁ = ½ 37.7 (1/3)²

         θ₁ = 2.09 rad

The angle for a full circle is θ = 2π rad, so the percentage is

        θ% = $\frac{\theta_1}{2\pi } \ 100$  

        θ%= $\frac{2.09}{2\pi} \ 100$  

        % = 33.2%

In conclusion, using the angular kinematic relations we can find the results on the motion questions are:

       a) The angular acceleration is:     α = 137.7 rad/s²

      b) The tangential acceleration is: a = 41.5 m/s²

      c) the maximum tangential speed is: v= 13.8 m/s

      d) The fraction of angle rotated is  θ% = 33.2%

Learn more about rotational kinematics here: brainly.com/question/20319147