A uniform disk with radius R = 0.400 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 θ(t) = (1.10 rad/s)t + (6.30 rad/s2)t. 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?

This question involves the concepts of the angular displacement and angular acceleration.

The resultant linear acceleration of a point on the rim at that instant will be "5.04 m/s²".

The angular velocity can be obtained by taking derivative of angular displacement with respect to time:

The angular acceleration can be obtained by taking derivative of angular displacement with respect to time:

Hence, the linear acceleration at any instant can be calculated using the following formula:

a = 5.04 m/s²

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The angle of rotation given is incorrect, the correct function to have a balanced unit is

θ(t) = (1.10 rad/s)t + (6.30 rad/s²)t²

Explanation:

Given that,

Radius of disk is

R = 0.4m

Mass of disk

M = 30kg

The angle of rotation is a function of time and it is given as

 θ(t) = (1.10 rad/s)t + (6.30 rad/s²)t²

What is linear angular acceleration "a" when it rotates 0.1rev

θ = 0.1 rev

1 rev = 2πrad

θ  = 0.1 × 2πrad = 0.63 rad

Let calculate the time it revolves 0.63rad

θ(t) = (1.10 rad/s)t + (6.30 rad/s²)t².

0.63 = 1.1t + 6.3t²

6.3t² + 1.1t - 0.63 = 0

Solving this quadratic equation using formula method

t = [-b±√(b²-4ac)]/2a

a = 6.3 b = 1.1 c = -0.63

t = [-1.1 ± √(1.1²-4×6.3×-0.63)]/2×6.3

t = -1.1±√(1.21+15.876) / 12.6

t = -1.1 ± 4.13 / 12.6

Let discard the negative time

So, t = (-1.1 + 4.13) / 12.6

t = 0.24 s

Now,

Angular velocity can be determine using

ωz = dθ / dt

ωz = 1.1 + 12.6t, since t = 0.24

ωz = 1.1 + 12.6(0.24)

ωz = 4.124 rad/s

Angular acceleration can be determine using

αz = dωz/dt

αz = 12.6 rad/s²

The radial acceleration can be determined using

ar = ωz²•R

ar = 4.124² × 0.4

ar = 6.8 m/s²

Also, the tangential Acceleration is related to angular acceleration using

at = αz • R

at = 12.6 × 0.4

at = 5.04 m/s²

Then, the magnitude of the resultant acceleration can be calculated using

a = √(ar² + at²)

a = √(6.8² + 5.04²)

a = √71.64

a = 8.46 m/s²

Direction

β = Arctan(at/ar)

β = Arctan(5.04/6.8)

β = 36.54

This is the direction from the origin

But it has already revolve 0.1rev

1 rev = 360°

Then, θ = 0.1° = 360 × 0.1 = 36°

Direction = β — θ

Then, it direction is 36.54 — 54° = 0.54°

The direction is 0.54°

Check attachment for direction understanding

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