A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is rotating at 10.0 rev/s; 65.0 revolutions later, its angular speed is 19.0 rev/s. Calculate (a) the angular acceleration (rev/s2), (b) the time required to complete the 65.0 revolutions, (c) the time required to reach the 10.0 rev/s angular speed, and (d) the number of revolutions from rest until the time the disk reaches the 10.0 rev/s angular speed.

Explanation:

The disk rotates from rest

Then,

Initial angular velocity ωi = 0 rev/s

At one time it angular velocity

ωi = 10 rev/s

And it's angular displacement is

θ = 65 rev

And later after 65rev, its angular velocity is 19rev/s

ωf = 19rev/s

A. Angular acceleration α?

Using the equation of circular motion

ωf² = ωi² + 2αθ

19² = 10² + 2 × α × 65

361 = 100 + 130α

361 — 100 = 130α

130α = 261

α = 261 / 130

α = 2.01 rev/s²

The angular acceleration is 2.01 rev/s²

B. The time required to complete the 65 revolution

Using a circular motion equation

ωf = ωi + αt

19 = 10 + 2.10t

19 — 10 = 2.10t

2.01t = 9

t = 9 / 2.01

t = 4.48 seconds

C. Time required to reach 10rev/sec

We know that initial the disk starts from rest

Now, ωi = 0 rev/s

The final angular velocity for this case is ωi = 10rev /s

Then, applying circular motion equation

ωf = ωi + αt

10 = 0 + 2.10t

10 = 2.10t

2.01t = 10

t = 10 / 2.01

t = 4.98 seconds

D. Number of revolution until it reaches 10 rev/s

Using equation of circular motion

θ = ωi•t + ½αt²

The initial angular acceleration is zero and the time to get to 10rev/s, has been calculated in C above

t= 4.98s

θ = ωi•t + ½αt²

θ = 0•t + ½× 2.10 × 4.98²

θ = = 0 + 25.99

θ = 25.99 revolutions

θ ≈ 26 revs

answer: the rock sample is a sedimentary rock.

explanation:

it is given that:

mass of rock sample, m = 2.7 g

volume of rock sample,

so, density of rock sample will be

since, the density of rock sample is . hence, the rock sample is a sedimentary rock, whose densities varies between .