Question

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.

Answer 1

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`