Question

This figure (∣

a
∣=18.5 m/s
2
) represents the total acceleration of a particle moving clockwise in a circle of radius f=2.40 m at a certain instant of time. (C) (a) For that instant, find the radial acceleration of the particle: m/s
2
(toward the center) (b) For that instant, find the speed of the particle. mis (c) For that instant, find its tangential acceleration, m/s
2
(in the direction of the motion)

Answer 1

(a) The radial acceleration of the particle is
`\( 18.5 \, \text{m/s}^2 \)` toward the center.

(b) The speed of the particle at that instant is
`\( 6.45 \, \text{m/s} \)`.

(c) The tangential acceleration of the particle is
`\( 0 \, \text{m/s}^2 \)` (since it's moving at constant speed).

Step-by-step explanation:

(a) The radial acceleration
`(\( a_r \))` can be determined using the formula
`\( a_r = (v^2)/(r) \)`, where
`\( v \)` is the speed and
`\( r \)` is the radius of the circle. Given that
`\( |a| = 18.5 \, \text{m/s}^2 \)` represents the total acceleration, and the particle is moving in a circle, the radial acceleration is
`\( 18.5 \, \text{m/s}^2 \).`

(b) The speed
`(\( v \))` of the particle can be found using the formula
`\( |a| = (v^2)/(r) \).` Rearranging for
`\( v \),` we get
`\( v = √( |a| \cdot r) \),` where
`\( r \)` is the radius. Substituting the given values, we find
`\( v \approx 6.45 \, \text{m/s} \).`

(c) Tangential acceleration
`(\( a_t \))` is the component of acceleration in the direction of motion. Since the particle is moving at a constant speed in a circular path, there is no change in speed, and
`\( a_t = 0 \, \text{m/s}^2 \).`