Question

A uniform disk with radius 0.310 m and mass 30.0 kg rotates in a horizontal plane on a frictionless vertical axle that passes through the center of the disk. The angle through which the disk has turned varies with time according to ?(t)=( 1.10 rad/s)t+( 6.90 rad/s2 )t2

What is the resultant linear acceleration of a point on the rim of the disk at the instant when the disk has turned through 0.100 rev ?

Answer 1

?(t) = 1.10t + 6.90t²

?(0) = 0

so the disk starts from rest

angular velocity is the first derivative of angular position

?'(t) = ?(t) = 1.10 + 13.8t rad/s

angular acceleration is the first derivative of angular velocity or the second derivative of angular position

?'(t) = ?'(t) = ?(t) = 13.8 rad/s²

as the angular acceleration is constant the tangential acceleration will be constant as well

at = ?r
at = 13.8(0.310)
at = 4.278 m/s²

to find the centripetal acceleration, we need to find the angular velocity when the disk has turned 0.100 rev

?(t) = 0.100(2?)
?(t) = 0.628 rad

0.628 = 1.10t + 6.90t²
0 = 6.90t² + 1.10t - 0.628

quadratic formula

t = (-1.10 ±?(-1.10² - 4(6.90)(-0.628))) / (2(6.90))

t = 0.2112 s
or
t = -0.3706 s which we ignore as it occurs before the disk starts spinning at t = 0 s

angular velocity at 0.2112 s

? = 1.10 + 13.8(0.2112)
? = 4.014 rad/s

centripetal acceleration is

ac = ?²r
ac = 4.014²(0.310)
ac = 4.994 m/s²

total linear acceleration is the vector sum of centripetal and tangential accelerations.

a = ?(4.278² + 4.994²)
a = 6.575 m/s²