Number of results: 104,151
After falling a distance h, Potential energy m g h is converted to kinetic energy of the rotating shell, the falling mass, and the pulley. Let V be the velcoity of the mass at time t. The angular velocity of the shell is ws= V/R and its moment of inertia is Is = (2/3)MR^2 The ...
Sunday, January 27, 2008 at 5:18pm by drwls
Use the law of conservation of angular momentum, measured about the vertical axis. Initially, the rod has no angular momentum but the bullet has angular momentum m V (L/2) cos 60 about that axis. Afterwards, the rod with embedded bullet has moment of inertia I = (1/12)ML^2 + m...
Friday, February 8, 2008 at 6:10pm by drwls
Add the gravitational forces due to the two spheres at the oppostice corners. The resultant force will be in a direction between them (30 degrees from each). Get the gravitational force between each pair using Newon's universal equation of gravity, F = G M1 M2/R^2 The ...
Thursday, March 6, 2008 at 8:26am by drwls
A uniform plate of height = 1.80 m is cut in the form of a parabolic section. The lower boundary of the plate is defined by: = 1.20 . The plate has a mass of 1.29 kg. Find the moment of inertia of the plate about the y-axis.
Wednesday, March 19, 2008 at 10:32pm by shelby
You need to clarify this statement: <The lower boundary of the plate is defined by: = 1.20 . > and somehow establish the width X of the parabola at the open end. The moment of inertia will be some fraction of M X^2 that can be established by calculus.
Wednesday, March 19, 2008 at 10:32pm by drwls
A uniform plate of height H= 0.69 m is cut in the form of a parabolic section. The lower boundary of the plate is defined by: y= 1.10x^2. The plate has a mass of 5.62 kg. Find the moment of inertia of the plate about the y-axis. Alright, I've done this so far: So I figured I ...
Friday, March 21, 2008 at 10:04pm by Kara
The half-width of the parabolic section at any height y above the x axis is x = sqrt (y/1.1) Perform one integral for the total mass in terms of H, using the known mass. This will tell you the density x thickness. Then perform a second integral for the moment of inertia, using...
Friday, March 21, 2008 at 10:04pm by drwls
questions( there are 3 so I will # 1-3) info: A 18.9 kg object is attached to a cord that is wrapped around a wheel of radius 10.7 cm. The acceleration of the object down the frictionless incline is measured to be .9 m/s2. Assume the axle of the wheel to be frictionless. The ...
Monday, March 24, 2008 at 4:18pm by Ty
Start with the tension. T= mg-ma Then the moment of inertia... Torque=I *angular acceleration Force*radius= I * linear acceleration/rad solve for I.
Monday, March 24, 2008 at 4:18pm by bobpursley
For ice, you can use conservation of energy. If h is the height that the block descends, (1/2)MV^2 = M g h V = sqrt (2gH) For the marble block, since it rolls, I assume it is a cylinder. Static friction is what helps it roll instead of slip, but it does not result in a of of ...
Sunday, April 6, 2008 at 10:56pm by drwls
A curcial part of a piece of machinery starts as a flat uniform cylindrical disk of radius R0 and mass M. It then has a circular hole of radius R1 drilled into it. The hole's center is a distance h from the center of the disk. Find the moment of inertia of this disk (with off-...
Saturday, April 12, 2008 at 4:13am by Sleepless
The moment of inertia of the disc before the hole is drilled is I0 = (1/2) M R0^2. From that, SUBTRACT the moment of inertia of what the hole would be, if it were solid and displaced h from the center. Its mass would be m = M(R1/R2)^2 Using the parallel axis theorem, its ...
Saturday, April 12, 2008 at 4:13am by drwls
The outstretched hands and arms of a figure skater preparing for a spin can be considered a slender rod pivoting about an axis through its center. When the skater's hands and arms are brought in and wrapped around his body to execute the spin, the hands and arms can be ...
Saturday, April 12, 2008 at 4:52pm by Erin
The total moment of inertia with arms outstretched is: I1 = Ibody + Iarms = 0.450 + (1/12) m L^2 = 0.450 + (1/12)*8.5*(1.8)^2 = 2.75 kg m^2 The total moment of inertia with arms rapped aound the body is: I2 = Ibody + Iarms = 0.450 + m R^2 = 0.45 + 0.53 = 0.98 kg m^2 Use the ...
Saturday, April 12, 2008 at 4:52pm by drwls
M = 110 m = 50 r = 4 The moment of inertia of the disk about the axis is .5 M r^2 The moment of inertia of the woman about the axis is m r^2 Total moment of inertia I = .5 Mr^2 + m r^2 angular velocity w = 2 pi * .5 radians/second angular momentum = I w = [.5 M + m] r^2* (2 pi...
Saturday, April 12, 2008 at 8:48pm by Damon
Since the pulled string exerts the only force on the disc (assuming that gravity is ignored here), you can use F = m a for the acceleration of the center of mass, in both cases. Pulling the string also starts the disk spinning at an acclerating spin rate. Use the relation ...
Saturday, April 12, 2008 at 10:18pm by drwls
You are probably suppoased to assume that the turntable is frictionless and "coasting" (unpowered) while the dry ice evaporates. This is a rather unrealistic assumption Anyway, in that case, you should assume that angular momentum is conserved. I1 w1 = I2 w2. If the dry ice is...
Monday, April 14, 2008 at 6:00pm by drwls
You are probably supposed to assume that the turntable is frictionless and "coasting" (unpowered) while the dry ice evaporates. This is a rather unrealistic assumption Anyway, in that case, you should assume that angular momentum is conserved. I1 w1 = I2 w2. If the dry ice is ...
Monday, April 14, 2008 at 6:00pm by drwls
If you model the turntable-ice mixture, then the moment of inertia will be proportional to mass. wf=Iinitial/I final* wi =k(4.08+22.8)/k(22.8) * .52 rad/sec
Monday, April 14, 2008 at 6:00pm by bobpursley
How do you find the moment of inertia of an actual bicycle about an axis through the pedal crank?
Thursday, October 2, 2008 at 10:13am by Sivaji