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.

1 Answer
Nov 21, 2017

Centripetal Acceleration: a_c = \sqrt{a_x^2 + a_y^2} = 30.24 ms^{-2}
Speed & Velocity: \qquad v = \sqrt{ra_c} = 6.73 ms^{-1}
vec v = v\cos\theta hat x + v\sin\theta hat y = (4.04 ms^{-1}) hat x + (5.38 ms^{-1}) 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 ms^{-2}

a_c = \sqrt{a_x^2 + a_y^2} = \sqrt{(-22.5)^2 + (20.2)^2} ms^{-2} = 30.24 ms^{-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 = \sqrt{ra_c} = \sqrt{(1.5 m)\times(30.24 ms^{-2})} = 6.73 ms^{-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 \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\cos\theta hat x + v\sin\theta hat y
vec v = (4.04 ms^{-1}) hat x + (5.38 ms^{-1}) hat y