Question

A cat rides a merry-go-round while turning with uniform circular motion. At time t1 = 2.00 s, the cat's velocity is v with arrow1 = (1.70)i hat + (4.00 m/s)j, measured on a horizontal xy coordinate system. At time t2 = 4.00 s, a half-revolution later, its velocity is v with arrow2 = (-1.70 m/s)i hat + (-4.00 m/s)j. (a) What is the magnitude of the cat's centripetal acceleration?
m/s2

(b) What is the magnitude of the cat's average acceleration during the time interval t2 - t1?
m/s2

Answer 1

Answer:

(a)

$a_c=3.41\frac{m}{s^2}$

(b)

$a_a_v_g=0\frac{m}{s^2}$

Explanation:

Let's calculate the magnitude of cat's velocity at time t1 and time t2:

$At\hspace{3}t_1$

$|v_1|=\sqrt{(v_x)^{2}+(v_y)^{2}  } =\sqrt{(1.70)^{2}+(4)^{2}  } =4.346262762m/s$

$At\hspace{3}t_2$

$|v_2|=\sqrt{(v_x)^{2}+(v_y)^{2}  } =\sqrt{(-1.70)^{2}+(-4)^{2}  } =4.346262762m/s$

For now:

$|v_1|=|v_2|$

So, we can assume that cat's tangential velocity is constant. Now, considering that the time T required for one complete revolution is called the period. For  constant speed is given by:

$T=\frac{2*\pi *r}{v}$  (1)

Where:

$r=radius\hspace{3}of\hspace{3}curvatrue$

$v=tangential\hspace{3}velocity$

The problem tell us that at time t2=4 the cat has completed a half-revolution, so we can conclude that the cat complete a revolution in 8s, T=8s. Replacing the data in (1) and isolating r:

$r=\frac{T*v}{2*\pi} =\frac{8*4.346262762}{2*\pi}=5.533833621m$

Now the centripetal acceleration is given by:

$a_c=\frac{v^{2} }{r} =\frac{(4.346262762)^{2} }{5.53383621} =3.413546791m/s^2$

Finally, The average acceleration is the final velocity minus the initial velocity per time taken.

$a_a_v_g=\frac{\Delta v}{\Delta t} =\frac{v_2-v_1}{t_2-t_1} =\frac{4.346262762-4.346262762}{4-2}=\frac{0}{2} =0$