# A question on circular motion?

## A ball swings in a vertical circle at the end of a rope 1.50 m long. When the ball is 36.9 degrees past the lowest point on its way up, its total acceleration is (−22.5+ 20.2j)m/s^2. At that instant, (a) sketch a vector diagram showing the components of its acceleration. (b)determine the magnitude of its radial acceleration. (c) determine the speed and velocity of the ball.

Nov 21, 2017

Centripetal Acceleration: ${a}_{c} = \setminus \sqrt{{a}_{x}^{2} + {a}_{y}^{2}} = 30.24$ $m {s}^{- 2}$
Speed & Velocity: $\setminus q \quad v = \setminus \sqrt{r {a}_{c}} = 6.73$ $m {s}^{- 1}$
$\vec{v} = v \setminus \cos \setminus \theta$ $\hat{x} + v \setminus \sin \setminus \theta$ $\hat{y} = \left(4.04 m {s}^{- 1}\right)$ $\hat{x} + \left(5.38 m {s}^{- 1}\right)$ $\hat{y}$

#### Explanation:

(B) Centripetal Acceleration:
vec a_c = a_x hat x + a_y hat y; \qquad a_x = -22.5 ms^{-2}; \qquad a_y = 20.2 $m {s}^{- 2}$

${a}_{c} = \setminus \sqrt{{a}_{x}^{2} + {a}_{y}^{2}} = \setminus \sqrt{{\left(- 22.5\right)}^{2} + {\left(20.2\right)}^{2}}$ $m {s}^{- 2} = 30.24$ $m {s}^{- 2}$

(C) Speed & Velocity: If the object is moving with a uniform speed $v$ along a circular path of radius $r$, its centripetal acceleration is,

a_c = v^2/r;
$v = \setminus \sqrt{r {a}_{c}} = \setminus \sqrt{\left(1.5 m\right) \setminus \times \left(30.24 m {s}^{- 2}\right)} = 6.73$ $m {s}^{- 1}$
This is the speed (magnitude of velocity vector).

The velocity vector is perpendicular to the acceleration vector. Since the velocity vector is tangential to the circle. When the ball is ${36.9}^{o}$ past the lowest point, the velocity vector is oriented at an angle of $\setminus \theta = {90}^{o} - {36.9}^{o} = {53.1}^{o}$ to the positive X-axis.

$\vec{v} = {v}_{x}$ $\hat{x} + {v}_{y}$ $\hat{y} = v \setminus \cos \setminus \theta$ $\hat{x} + v \setminus \sin \setminus \theta$ $\hat{y}$
$\vec{v} = \left(4.04 m {s}^{- 1}\right)$ $\hat{x} + \left(5.38 m {s}^{- 1}\right)$ $\hat{y}$