Question

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?

Answer 1

Answer:

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

Answer 2

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:

$\frac{d\theta}{dt}=\omega=\frac{d}{dt}(1.1t+6.3t^2)\\\\\omega = 12.6t+1.1$

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

$\frac{d\omega}{dt}=\alpha=\frac{d}{dt}(12.6t+1.1)\\\\\alpha = 12.6$

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

$a=R\alpha\\a=(0.4\ m)(12.6\ rad/s^2)$

a = 5.04 m/s²

Learn more about angular acceleration here:

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