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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.
b). an axis through C perpendicular to the page
magnitude ________ N*m
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
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
_________ kg * m^{2}
ii. hoop > thin spherical > solid cylinder > Solid sphere
iii. hoop > Solid Cylinder > solid Sphere> Thin spherical
iv. thin spherical > solid sphere > solid cylinder > hoop
b) a solid cylinder
__________ kg * m^{2}/s
__________ J * s
a) Find the new angular speed of the student
BONUS QUESTION:
+DrMohammed Salah ElGazzar 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:
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 33.5kg 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 upwardb) 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 49.7cm mark when placed on a fulcrum. When a 61.0gram mass is attached at the 12.0 cm mark, the fulcrum must be moved to the 39.2cm mark for balance. What is the mass of the meter stick?
__________ g
6. A 520N uniform rectangular sign 4.00 m wide and 3.00 m high is suspended from a horizontal, 6.00mlong, uniform, 140N 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 6.00mlong 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
_______ 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 710N worker stands 1.90 m from one end?
Smaller tension ________N
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 xaxis
_________kg*m^{2}
b) Find the moment of inertia of the system about the yaxis
_________kg*m^{2}
c) Find the moment of inertia of the system about an axis through O and perpendicular to the page.
________kg*m^{2}
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 * m^{2}
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 245kg merrygoround 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 merrygoround 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 810N merrygoround of radius 1.70 m is started from rest by a constant horizontal force of 55 N applied tangentially to the merrygoround. Find the kinetic energy of the merrygoround after 2.0 s. (assume it is a solid cylinder)
___________ J
13. Four objectsa hoop, solid cylinder, a solid sphere, and a thin, spherical shelleach 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 * m^{2}
Solid cylinder ________ kg * m^{2}
Solid sphere ________ kg * m^{2}
Thin, spherical shell ________ kg * m^{2}
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> hoopii. 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 m_{1} = 4.45 kg and m_{2} = 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 240N 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 * m^{2}/sb) a solid cylinder
__________ kg * m^{2}/s
c) a solid sphere
__________ kg * m^{2}/s
d) a hollow spherical shell
_______ kg * m^{2}/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 m_{1} = 4.55 kg and m_{2} = 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 2.8kg 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 * m^{2} 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:
+DrMohammed Salah ElGazzar 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 WorkEnergy 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 copycat.
 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 shelllike 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.
Thanks for the help! Much appreciated :)
ReplyDeleteAlways glad to help! Are you taking physics at MWSU?
ReplyDeleteyou saved my life! Thank you so much :) So useful with my midterm coming up in two days.
ReplyDeleteStudy hard! Physics II videos (Chapter 15  End of book) will be posted in the spring. Spread the word!
DeleteThanks for the help! I was looking for the link you were supposed to upload for problem #4 but, I didn't see it.
ReplyDeleteI posted it for you. http://www.youtube.com/watch?v=Y3kcxu3Qr8E
DeleteSir, you have helped out so much during the course of my Physics class. I just want to thank you for being so explanative and clear.
ReplyDeleteI think I was anything except for clear, but I'm really glad it helped.
DeleteI did this because I saw a need. As you can see from the site name, I originally wanted to focus solely on biology, but it seems there is a larger need in physics at this point in time.
I started taking Physics online and realized that I was going to have to work twice as hard as anyone who took the course in a classroom. I also had friends who were asking me for help, and I didn't have time to sit down with each one of them. So, these physics videos are dedicated to my friends
And anyone who finds these videos useful, I also consider an extended friend.
So, friend, I'm glad this has been helpful
Hey could you help with
ReplyDelete"When a person stands on tiptoe (a strenuous position), the position of the foot is as shown in figure (a). The total gravitational force on the body, vector F g, is supported by the force narrowbold exerted by the floor on the toes of one foot. A mechanical model of the situation is shown in figure (b), where vector T is the force exerted by the Achilles tendon on the foot and vector R is the force exerted by the tibia on the foot. Find the values of T, R, and θ when Fg = n = 740 N. (Do not assume that vector R is parallel to vector T .) "
it is one of my physics chapter 8 homework. I have no clue how to approach it.
send me the picture in question and I'll try to give you a good answer
DeleteI figured out how to do it but thanks anyways. Your website help me survive through physics! Great job and thank you for taking the time to do this.
DeleteHello Mike,
ReplyDeleteI have been getting help for my homework from your website. You are a life saver. Please, kindly help me out with the below question. I keep getting it wrong. I will greatly appreciate it. Thanks. It is from webassign.
The puck in Figure P8.51 has a mass of 0.140 kg. Its original distance from the center of rotation is 40.0 cm, and it moves with a speed of 80.0 cm/s. The string is pulled downward 15.0 cm through the hole in the frictionless table. Determine the work done on the puck. (Hint: Consider the change of kinetic energy of the puck.)
J
Figure P8.51
I also need help with this problem!
DeleteAnonymous:
DeleteWithout seeing figure P8.51, it would be very difficult for me to determine exactly what is being asked. But I can give you this general advice:
1. Remember the big idea here is workenergy theorom
2. Work energy theorom problems can be solved by replacing the following variables:
W = dKe + dPe + dKre (where d is delta/change, Ke is kinetic energy, Pe is potential energy, and Kre is rotational kinetic energy).
 You can expand this equation to include any kind of change in energy whether it be kinetic, potential, or heat transfers, but that is beyond the scope of your problem
3. Each variable will have an associated equation (dKe = 1/2 m(dv)^2 (where dv is the change in velocity). and Potential energy changes can be calculated by mg(d)h (change of height). Rotational kinetic energy changes will have equations that depend on the rotational moment of inertia (usually represented as a capital "I"), and the shape of the object impacts the equation
4. Remember, the equation is "Adding" all of the changes. Some of the changes will be "negative" so keep track of what's happening  because, instead of adding, you'll subtract any negative changes.
5. Post a picture (or a link to a picture) and I'll see if there is anything else I can do to help you out.
This comment has been removed by the author.
ReplyDeleteA 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.
ReplyDeleteRepeat this example for a solid cylinder of the same mass and radius as the ball and released from the same height.
+DrMohammed Salah ElGazzar,
DeleteI Typed an answer at the bottom of this blog entry (see above  after the very last question)
I definitely hope it helps!
How would you approach a problem with an initial angular speed of a solid uniform disk (you know M, R, rotating CW) and a initial (non angular) speed of a child running from its right and jumping on to the disk (you know childs m). How would you find the new angular speed for the disk with the child on its edge?
ReplyDeleteDo you have to convert the childs speed to an angular speed?