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 = (2.30)i hat + (4.00 m/s)j, measured on a horizontal xy coordinate system. At time t2 = 9.00 s, a half-revolution later, its velocity is v with arrow2 = (-2.30 m/s)i hat + (-4.00 m/s)j.

Answer 1

Answer:

Part a)

$a_c = 2.07 m/s^2$

Part b)

$a_{avg} = 1.32 m/s^2$

Explanation:

As we know that it makes half revolution in given time interval

so we have

$\frac{T}{2} = t_2 - t_1$

$\frac{T}{2} = 9 - 2$

$T = 14 s$

now the angular speed is given as

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{14}$

$\omega = 0.448 rad/s$

now linear speed is given as

$v = \sqrt{2.30^2 + 4.00^2}$

$v = 4.61 m/s$

now we have

$v = R \omega$

$4.61 = R(0.448)$

$R = 10.3 m$

Now centripetal acceleration is given as

$a_c = \omega^2 R$

$a_c = 0.448^2 \times 10.3$

$a_c = 2.07 m/s^2$

Part b)

Average acceleration of the cat is given as

$a_{avg} = \frac{v_2 - v_1}{\Delta t}$

$a_{avg} = \frac{2v}{\Delta t}$

$a_{avg} = \frac{2(4.61)}{9 - 2}$

$a_{avg} = 1.32 m/s^2$