A disk of radius 2.83 m rotates about its axis. Points on the disk's rim undergo tangential acceleration of magnitude 2.53 m/s2. At a particular time the rim has a tangential speed of 1.53 m/s. At a time 0.817 seconds later, what is the tangential speed, v, of a point on the rim, the magnitude of the point's radial acceleration, ar, and the magnitude of its total acceleration, atot?

find:

V=

a(radial)=

a(total)=

Explanation:

v = u + at

= 1.53 + 2.53 x0.817

= 3.6 m/s

Radial acceleration = v² / r

= 3.6 x 3.6 / 2.83

a_r = 4.58 m / s²

a_tot = √ 4.58² + 2.53²

= √ (20.97 + 6.4)

= 5.23 m /s²

(a). The tangential speed is 3.597 m/s.

(b). The magnitude of the point's radial acceleration is 4.57 m/s².

(c). The magnitude of its total acceleration is 5.22 m/s².

Explanation:

Given that,

Radius of disk = 2.83 m

Acceleration = 2.53 m/s²

Tangential speed = 1.53 m/s

Time = 0.817 s

(a). We need to calculate the tangential speed

Using equation of motion

Where, v = final velocity

a = acceleration

T = time

Put the value into the formula

(b). We need to calculate the magnitude of the point's radial acceleration

Using formula of radial acceleration

Put the value into the formula

(c). We need to calculate the magnitude of its total acceleration

Hence, (a). The tangential speed is 3.597 m/s.

(b). The magnitude of the point's radial acceleration is 4.57 m/s².

(c). The magnitude of its total acceleration is 5.22 m/s².