Physics Homework Ch8

Don't forget to +1, Like the videos, and share on Facebook


1. The fishing pole in the figure below makes an angle of 20.0° with the horizontal. What is hte magnitude of the torque exerted by the fish about the axis perpendicular to the page and passing through the angler's hand if the fish pulls on the fishing line with a force  = 103N at an angle 37.0° below the horizontal? The force is applied at a point L = 1.86 m from the angler's hands.

_________ N*m

2. Find the net torque on the wheel in the figure below about the axle through O perpendicular to the page, taking a = 7.00 cm and b = 21.0 cm (Indicate the direction with the sign of your answer. Assume that the positive direction is counterclockwise.)

________N*m

3. Calculate the net torque (magnitude and direction) on the beam in the figure below about the following axes.

a) an axis through O perpendicular to the page

magnitude ________ N*m

Direction: (Counterclockwise or Clockwise)

b). an axis through C perpendicular to the page
magnitude ________ N*m

Direction (Counterclockwise or Clockwise)

4. A uniform 33.5-kg beam of length l = 4.95 m is supported by a vertical rope located d = 1.20 m from its left end as in the figure below. The right end of the beam is supported by a vertical column.

a) Find the tension in the rope

__________ N upward

b) Find the force that the column exerts on the right end of the beam.
__________ N upward






http://www.youtube.com/watch?v=Y3kcxu3Qr8E
A "different" way to solve for Torque

5. A meter stick is found to balance at the 49.7-cm mark when placed on a fulcrum. When a 61.0-gram mass is attached at the 12.0 cm mark, the fulcrum must be moved to the 39.2-cm mark for balance. What is the mass of the meter stick?

__________ g

6. A 520-N uniform rectangular sign 4.00 m wide and 3.00 m high is suspended from a horizontal, 6.00-m-long, uniform, 140-N rod as indicated in the figure below. The left end of the rod is supported by a hinge and the right end is supported by a thin cable making a 30.0° angle with the vertical. (Assume the cable is connected to the very end of the 6.00-m-long rod, and that there are 2.00 m separating the wall from the sign.)

a) Find the (magnitude of the ) tension T in the cable.
_______ N 

b) Find the horizontal and vertical components of the force exerted on the left end of the rod by the hinge. (Take up and to the right to be the positive directions. Indicate the direction with the sign of your answer).
Horizontal Component ___________ N

Vertical Component ___________ N

7. A window washer is standing on a scaffold supported by a vertical rope at each end. The scaffold weighs 204 N and is 2.9 m long. What is the tension in each rope when the 710-N worker stands 1.90 m from one end?
Smaller tension ________N

Larger tension  ________N

8. Four objects are held in position at the corners of a rectangle by light rods as shown in the figure below. (The mass values are given in the table.)

m1 (kg)	m2 (kg)	m3 (kg)	m4 (kg)
2.7	1.7	3.5	1.5

a) Find the moment of inertia of the system about the x-axis

_________kg*m²

b) Find the moment of inertia of the system about the y-axis

_________kg*m²

c) Find the moment of inertia of the system about an axis through O and perpendicular to the page.

________kg*m²

9. If the system shown in the figure below is set in rotation about each of the axes listed below, find the torque that will produce an angular acceleration of 3.8 rad/² in each case.

x axis                                                           ________ N * m

y axis                                                           ________ N * m

axis through O and perpendicular to the page   ________ N * m

10. A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is free to rotate on a frictionless, vertical axle. A constant tangential force of 230 N applied to its edge causes the wheel to have an angular acceleration of 0.872 rad/².

a) What is the moment of inertia of the wheel?
_________ kg * m²

b) What is the mass of the wheel?

_________ kg

c) If the wheeel starts from rest, what is its angular velocity after 4.30 s have elapsed, assuming the force is acting during that time?

________ rad/s

11. A 245-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.700 rev/s in 2.00 s? (State the magnitude of the force.)

__________N

12. A horizontal 810-N merry-go-round of radius 1.70 m is started from rest by a constant horizontal force of 55 N applied tangentially to the merry-go-round. Find the kinetic energy of the merry-go-round after 2.0 s. (assume it is a solid cylinder)

___________ J

13. Four objects--a hoop, solid cylinder, a solid sphere, and a thin, spherical shell--each has a mass of 5.39 kg and a radius of 0.252 m

a) Find the moment of inertia for each object as it rotates about the axes shown in the table above.

hoop                         ________ kg * m²

Solid cylinder             ________ kg * m²

Solid sphere               ________ kg * m²

Thin, spherical shell    ________ kg * m²

b) Suppose each object is rolled down a ramp. Rank the translational speed of each object from highest to lowest.

c) Rank the objects's rotational kinetic energies from highest to lowest as the objects roll down the ramp.

i. Solid cylinder> thin spherical> Solid Cylinder> Solid sphere> hoop
ii. hoop > thin spherical > solid cylinder > Solid sphere
iii. hoop > Solid Cylinder > solid Sphere> Thin spherical
iv. thin spherical > solid sphere > solid cylinder > hoop

14. A light rod of length  l = 1.00 m rotates about an axis perpendicular to its length and passing through its center as in the figure below. Two particles of masses m₁ = 4.45 kg and m₂ = 3.00 kg are connected to the ends of the rod.

a) Neglecting the mass of the rod, what is the system's kinetic energy when its angular speed is 2.7 rad/s?

___________ J

b) Repeat the problem, assuming the mass of the rod is taken to be 1.60 kg.

__________ J

15. A 240-N sphere 0.20 m in radius rolls without slipping 6.0 m down a ramp that is inclined at 37° with the horizontal. What is the angular speed of the sphere at the bottom of the slope if it starts from rest?

_________ rad/s

16. Each of the following objects has a radius of 0.184 m and a mass of 2.50 kg, and each rotates about an axis through its center (as in the table below) with an angular speed of 35.9 rad/s. Find the magnitude of the angular momentum of each object

a) a hoop

_______ kg * m²/s
b) a solid cylinder
__________ kg * m²/s

c) a solid sphere

__________ kg * m²/s

d) a hollow spherical shell

_______ kg * m²/s

17.

a) Calculate the angular momentum of Earth that arises from its spinning motion on its axis, treating Earth as a uniform solid sphere.
__________ J * s

b) Calculate the angular momentum of Earth that arises from its orbital motion about the Sun, treating Earth as a point particle.

18. A light rigid rod of lenth l = 1.00 m in length rotates about an axis perpendicular to its length and through its center, as shown in the figure below. Two particles of masses m₁ = 4.55 kg and m₂ = 3.00 kg are connected to the ends of the rod. What is the angular momentum of the system if the speed of each particle is 6.40 m/s? (Neglect the rod's mass).

19. A student sits on a rotating stool holding two 2.8-kg objects. When his arms are extended horizontally, the objects are 1.0 m from the axis of rotation and he rotates with an angular speed of 0.75 rad/s. The moment of inertia of the student plus stool is 3.0 kg * m² and is assumed to be constant. The student then pulls in the objects horizontally to 0.28 m from the rotation axis.

a) Find the new angular speed of the student

_________ rad/s

b) Find the kinetic energy of the student before and after the objects are pulled in.

Before: __________ J

After  ___________ J

BONUS QUESTION:
+Dr-Mohammed Salah El-Gazzar asked me (in the comments below)
A ball of mass M and radius R starts from rest at a height of 2.00 m and rolls down a 30.0° slope, as in the figure. What is the linear speed of the ball when it leaves the incline? Assume that the ball rolls without slipping.

Repeat this example for a solid cylinder of the same mass and radius as the ball and released from the same height.


BONUS ANSWER:Here is your answer:

1st Principle:

The Work-Energy theorem:

Sum of all Changes of energy is equal to work

 -- If there is no friction, heat transfer, etc -- that is, if all forces acting are conservative forces, then the work will equal ZERO (0)

So:

Sum of changes in Linear kinetic Energy + Sum of changes in Rotational kinetic energy = - (negative) changes in potential energy

ΔKE(linear) + ΔKE(rotational) = ΔPE

ΔKE(linear) = (1/2)(m)(Δv^2)

2nd Principle:

For every linear variable, there is a rotational copy-cat. 

 - Mass, in linear kinematics is like "I" (represents moment of inertia --AKA rotational mass or rotational inertia--

 - Velocity, in linear kinematics is like "ω" (Omega represents angular velocity) 

ΔKE(rotational) = (1/2)(I)(Δω^2)

ω = velocity/radius

3rd Principle:

 the Moment of inertia is a quantity that describes the mass as it is distributed around the axis of rotation. 

 So, different geometries of mass will have slightly different equations for "I"

 -- In our case, we need to be able to calculate "I" for a Sphere and a Solid cylinder:

There is ONE big problem with your question at this point. 

 1. Is the "ball" a solid sphere (like a bowling ball) or a shell-like sphere (like a basket ball)?

I(solid sphere) = 2/5M(R^2)

I(hollow sphere) = 2/3M(R^2)

And Then finally

I(solid cylinder) = 1/2M(R^2)

So, let's do the equation for the solid cylinder, first, since we know exactly what formula to use:

--Start with the equation for the work energy theorem (shown in Eq 1, modified for all conservative forces).

Eq 1:

ΔKE(linear) + ΔKE(rotational) = -ΔPE

Eq 2/3/4

ΔKE(linear) = (1/2)(m)(Δv^2)

ΔKE(rotational) = (1/2)(I)(Δω^2)

ΔPE = mgΔh

Substitution of Eq 2/3/4 into Eq 1 gives the following: 

Eq 5

(1/2)(m)(Δv^2) + (1/2)(I)(Δω^2) = -mgΔh

Convert the rotational variables into 

Conversion 1

I = 1/2M(R^2)

Conversion 2

ω = v/r*

* NOTE: the use of capital "R" and lower case "r" both refer to the "radius", but convention has them being either capital or lower case in different conversions... They are the same  variable, here: r = R

Subsitute conversion 1 and conversion 2 into equation 5 provides:

Eq 6

(1/2)(m)(Δv^2) + (1/2)(1/2M(R^2))((Δv/r)^2) = -mgΔh

The mass can be divided out of this problem (all terms have mass)

Eq 7

(1/2)(Δv^2) + (1/2)(1/2(R^2))((Δv/r)^2) = -gΔh

This can be reduced via the following steps:

 - multiply both sides by "2"

 - Allow R^2 to cancel itself out

 - divide both sides by 1.5

Eq 8

Δv^2 = (-2/1.5)gΔh

Take the square root of both sides to solve for the change in linear velocity. Since we are assuming the ball started at rest (zero velocity), then the Δv is equal to the final velocity (when it leaves the incline)

v = Square root((-2/1.5)gΔh)

 -- Remember that the change in height (final height = 0, intial height was 2 meters) will be a negative number (final minus intial-- 0 - 2 = -2), so the negatives will cancel each other out.

Or

 v = Square root ((-1.333)*(9.8)*(-2)) = 26.13268 m/s (FOR THE CYLINDER)

Now, assume the ball is solid sphere (not hollow) -- everything starts the same:

Eq 1:

ΔKE(linear) + ΔKE(rotational) = -ΔPE

Eq 2/3/4

ΔKE(linear) = (1/2)(m)(Δv^2)

ΔKE(rotational) = (1/2)(I)(Δω^2)

ΔPE = mgΔh

Substitution of Eq 2/3/4 into Eq 1 gives the following: 

Eq 5

(1/2)(m)(Δv^2) + (1/2)(I)(Δω^2) = -mgΔh

HERE IS WHERE THINGS CHANGE -- "I" HAS A DIFFERENT CONVERSION EQUATION

Convert the rotational variables into 

Conversion 1

I = 2/5M(R^2)

Conversion 2

ω = v/r*

Subsitute conversion 1 and conversion 2 into equation 5 provides:

Eq 6

(1/2)(m)(Δv^2) + (1/2)(2/5M(R^2))((Δv/r)^2) = -mgΔh

- Again, divide out the "mass"

 - multiply both sides by "2"

 - Allow R^2 to cancel itself out

 - divide both sides by 1.5

 - take the square root of both sides

v = Square root( (10/7) X (9.8) X (2) )

v = 5.291502622129181 m/s (for a solid sphere)

By now, I think you probably understand what I'm doing -- just use a different conversion factor for 

"I" and all of the other steps will be the exact same.