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PHYS 1501Q Lecture Notes - Lecture 29: Angular Velocity, Angular Acceleration, Kinetic Energy

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Verified Note
coralcat765
1 Nov 2018
School
UConn
Department
Physics
Course
PHYS 1501Q
Professor
Soroush Khosravi Dehaghi, Robert McCutcheon

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PHYS 1501Q Lecture Notes - Lecture 28: Angular Acceleration, International System Of Units, If And Only If

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PHYS 1501Q Lecture Notes - Lecture 29: Angular Velocity, Angular Acceleration, Kinetic Energy

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PHYS 1501Q Lecture Notes - Lecture 30: Simple Harmonic Motion, Equilibrium Point, Kinematics

Document Summary

Physics 1501q - lecture 29 - conservation of angular momentum . Key (in si units): t = time (s) x = distance (m) v = velocity (m/s) a = acceleration (m/s 2 ) = change in g = gravity = 9. 81 m/s 2. G = 6. 67 10 11 n m 2 /kg 2 m = mass (kg) U = potential energy (j) w = angular velocity (m/s) A disk of mass m = 0. 40 kg and radius r = 50 cm rotates counterclockwise freely around its center with an angular velocity of w 0 = 5. 1 rad/sec, as shown. You toss a block of mass m = 0. 10 kg horizontally onto the disk along the direction shown at distance r/2 from the center. Friction acts between the disk and the block so the block comes to rest relative to the disk and starts rotating with it.

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Related textbook solutions

Physics: Principles with Applications
7th Edition, 2013
Giancoli
ISBN: 9780321625922

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Related Questions

1) You accelerate your car from rest at a constant rate downastraight road, and reach 22.0 m/s in 117 s. The tires on yourcarhave radius 0.320 m. Assuming the tires rotate inacounterclockwise direction, what is the angular acceleration ofthetires?

2) A pulley has aradius of 2.70 m and a moment of inertia of39.0 kg·m2. Thehanging mass is 4.20 kg and it exerts a forcetangent to the edgeof the pulley. What is the angular accelerationof thepulley?
3) A pulley is uniform disk of mass 2.8 kg and radius0.24 m,which is free to rotate without friction. The mass of theblock is1.2 kg. Downward and clockwise are negative directions.(a) What isthe angular acceleration of the pulley? (b) What is theaccelerationof the falling block?
4) A pulley Ca is a uniform disk of mass 2.40 kg andradius0.220 m. The block applies a contant torque to the pulley,which isfree to rotate without friction, resulting in anangularacceleration of magnitude 0.180 rad/s2 for the pulley. Asthe blockfalls 0.500 m, how much work does it do on thepulley?
A small piece of cheese is at the center of a disk of radiusRrotating about its axis. A hungry mouse is sitting at the edgeofthe disk. It will be able to eat the cheese when it is a distancer= R/4 from the center of the disk. The surface of thedisk,however, is slippery ...
In the following, assume that the disk is spinning freely aboutitsaxis and that there is no net external torque acting onthesystem.
a) Write expressions for the moment of inertia of the system(diskplus mouse) when the mouse is at the edge (distanceR) andwhen the mouse is at the distance r fromthe center. Themoment of inertia of the disk is I=(1/2)MR2.
b) As the mouse approaches the center, does the disk speed up,slowdown, or maintain its angular velocity?
c) Assuming the angular velocity is wwhen the mouse is at the edgeof the disk, find an expression forthe angular velocitywr whenthe mouse is a distance r = R/4from thecenter.
d) Evaluate the angular velocity wr for the case thatthe mass of thedisk is M = 18 g, the mass of the mouse ism = 16g, and w = 4 s-1. Answer:wr =10 s-1.
e) Determine if a static friction coefficient of ms =0.7 is sufficient to prevent themouse from slipping at r =R//4 for R =0.4 m. Please use g = 10m/s2.

Suppose there is a little block attached to a cord passing overapulley through a hole in the


show all calculations and work ...THANKS!

Suppose there is a little block attached to a cord passingovera pulley through a hole in the horizontal frictionlesssurface. Atfirst the little block is revolving with an initialangular speed of6rad/sec at a distance r from the center of thathole. So if thischord is pulled from below until the new radius isr/6, then whatwill be the new angular velocity?


Show all calculations and work ...THANKS!

w=work andw(i)=initial work
r= radius
m=mass

Hints given: L= r*p- = r*m *v(1)

v- =w*r

L(i)=r(i)*m*w(i)*r(i)

Eventually masses will cancel out so leftwithr(i)^2*w(i)--> solve for w(i)


L(f)=r(f)*m*w(f)*r(f)

Likewise mass will cancel out left r(f)^2*w(f)

Then plug in the values into equation L(i)=L(f) (setthem equalto each other to cancel out masses that way)

Conservation of angular momentum L(i)=L(f)
=L(0) -->the angular momentum constant

Solve forW(f)