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?

A)  = 37.68 rad/

B) a = 41.45 m/ = 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 = + 1/2            

As initial speed is zero so,

Ф1 = 0 + 1/2  

Ф1 = 0.05445     equation 1

And final speed will be:

w = +

w = 0 + (0.33)

w = 0.33    equation 2

Similarly, for second part of the circle.

Ф2 = w

(2 - Ф1) = (0.33) (t-)

(2 - 0.05445 ) = (0.33 ) (0.67 - 0.33)

Solving the above equation for , we get:

 = 37.68 rad/

B) For tangential acceleration, we know that:

a = r

r = 1.1 m

a = (1.1) (37.68)

a = 41.45 m/ = Tangential acceleration

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

v = rw

r = 1.1 m

w = 0.33

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

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 x 100

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

% age of full circle rotation = 32.64%

neither accurate nor precise.

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