Physics, 30.03.2020 17:45 Alexisgrab

A 2.00-kg textbook rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is 0.150 m, to a hanging book with mass 3.00 kg. The system is released from rest, and the books are observed to move 1.20 m in 0.800 s. (a) What is the tension in each part of the cord? (b) What is the moment of inertia of the pulley about its rotation axis?

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Answer from: shelbyhood8329

(a) The tension on the first book is 7.5 N and the tension of the second book is 18.15 N.

(b) The moment of inertia of the pulley about its rotation is 0.016 kgm².

The given parameters:

mass of the textbook, m = 2 kgdiameter of the cord, d = 0.15 cmmass of the hanging book, m = 3 kgdistance moved by the book, s = 1.2 mtime of motion, t = 0.8 s

The acceleration of the book is calculated as follows;

$s = v_0t + \frac{1}{2} at^2\\\\1.2 = 0 \ + \ \frac{1}{2} (a)(0.8)^2\\\\1.2 = 0.32a\\\\a = \frac{1.2}{0.32} \\\\a = 3.75 \ m/s^2$

The tension on the first book is calculated as follows;

$T_1 = m_1a \\\\T_1 = 2 \times 3.75\\\\T_1 = 7.5 \ N$

The tension on the second book is calculated as follows;

$T_2 = W_2 - m_2a\\\\T_2 = m_2 g - m_2 a\\\\T_2 = m_2(g-a)\\\\T_2 = 3(9.8 - 3.75)\\\\T_2 = 18.15 \ N$

The moment of inertia of the pulley about its rotation is calculated as follows;

$\Sigma \tau = I\alpha \\\\(T_2- T_1)r = I \times \frac{a}{r} \\\\(T_2- T_1)r^2 =I a\\\\I = \frac{(T_2- T_1)r^2}{a} \\\\I = \frac{(18.15 - 7.5) \times 0.075^2}{3.75} \\\\I =0.016 \ kg m^2$

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A 2.00-kg textbook rests on a frictionless, horizontal surface. A cord attached to the book passes o

Answer from: josiebranstetter93

Given that,

Mass of textbook Mb = 2kg

Initially at rest Ub=0m/s

diameter of pulley d = 0.15m

Then, the radius is r = d/2 =0.15/2 = 0.075m

A hang book of mass Mh = 3kg

Initial at rest too Uh = 0m/s

Distance moved = 1.2m in t = 0.8s

Check attachment for free body diagram and better understanding

Analysis of the mass of textbook on table

Using Newton second law

ΣFx = m•ax

T1 = Mb•ax

T1 = 2•ax, equation 1

Now, analysis of the hanged body

Using the same newton's law

ΣFy = m•ay

T2 — Mh•g = —3•ay,

ay is negative because it is moving in the negative direction

Then, T2 = Mh•g — 3•ay

T2 = 3×9.81 —3•ay

T2 = 29.43 — 3•ay, equation 2

Now, the body moves 1.2m in t=0.8second

The initial velocity of the body is 0m/s

Using equation of motion to calculate the acceleration (a)

S=ut+½at²

1.2 = 0•t + ½ × a × 0.8²

1.2 = 0 + 0.32a

0.32a =1.2

a = 1.2/0.32

a = 3.75m/s²

Since the body are connected by an I inextensive string, their acceleration are the same i.e. ax=ay=a=3.75m/s²

So, back to equation 1

T1 = 2•ax

T1 = 7.5 N

Also, back to equation 2

T2 = 29.43 — 3•ay

T2 = 29.43 — 3 × 3.75

T2 = 29.43 — 11.25

T2 = 18.18 N

b. Torque?

Modelling the pulley as a rigid body

Then, applying equilibrium of torque

Clockwise torque = anti-clockwise torque

Σ τ = (T2-T1)r = Iα

Where α is angular acceleration

Relationship between angular acceleration and radial acceleration is given as a=αr

Therefore,

(T2-T1)r = Iα

Since α=a/r

(T2-T1) = Ia/r

Cross multiply

(T2-T1)r² = Ia

(18.18 — 7.5) × 0.075² = I(3.75)

10.68×0.075² = 3.75I

0.060075 = 3.75I

Then, I = 0.060075/3.75

I = 0.01602 kgm²

I ≈ 0.016 kgm²

Then, the moment of inertia is 0.016 kgm²

Answer from: Quest

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2

Answer from: Quest

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A 2.00-kg textbook rests on a frictionless, horizontal surface. A cord attached to the book passes o...