Question

A) M, a solid cylinder (M=1.67 kg, R=0.127 m) pivots on a thin, fixed, frictionless bearing....

A) M, a solid cylinder (M=1.67 kg, R=0.127 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls downward with a force F which equals the weight of a 0.730 kg mass, i.e., F = 7.161 N. Calculate the angular acceleration of the cylinder.

B) If instead of the force F an actual mass m = 0.730 kg is hung from the string, find the angular acceleration of the cylinder.

C) How far does m travel downward between 0.690 s and 0.890 s after the motion begins?

D) The cylinder is changed to one with the same mass and radius, but a different moment of inertia. Starting from rest, the mass now moves a distance 0.337 m in a time of 0.450 s. Find Icm of the new cylinder.

Science   Physics   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

0 0
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
M, a solid cylinder (M=1.47 kg, R=0.135 m) pivots on a thin, fixed, frictionless bearing. A...
M, a solid cylinder (M=1.47 kg, R=0.135 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls downward with a force F which equals the weight of a 0.690 kg mass, i.e., F = 6.769 N. Calculate the angular acceleration of the cylinder. Tries 0/8 If instead of the force F an actual mass m = 0.690 kg is hung from the string, find the angular acceleration of the cylinder. Tries 0/8 How far does...
a) M, a solid cylinder (M=2.43 kg, R=0.137 m) pivots on a thin, fixed, frictionless bearing....
a) M, a solid cylinder (M=2.43 kg, R=0.137 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls downward with a force F which equals the weight of a 0.710 kg mass, i.e., F = 6.965 N. Calculate the angular acceleration of the cylinder. b)If instead of the force F an actual mass m = 0.710 kg is hung from the string, find the angular acceleration of the cylinder. c)How far does m travel downward...
M, a solid cylinder (M=1.39 kg, R=0.115 m) pivots on a thin, fixed, frictionless bearing. A...
M, a solid cylinder (M=1.39 kg, R=0.115 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls downward with a force F which equals the weight of a 0.830 kg mass, i.e., F = 8.142 N. 1.) Calculate the angular acceleration of the cylinder. 2.) If instead of the force F an actual mass m = 0.830 kg is hung from the string, find the angular acceleration of the cylinder. 3.) How far does m...
M, a solid cylinder (M=1.59 kg, R=0.111 m) pivots on a thin, fixed, frictionless bearing. A...
M, a solid cylinder (M=1.59 kg, R=0.111 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls downward with a force F which equals the weight of a 0.870 kg mass, i.e., F = 8.535 N. How far does m travel downward between 0.530 s and 0.730 s after the motion begins? The cylinder is changed to one with the same mass and radius, but a different moment of inertia. Starting from rest, the mass...
A small, solid cylinder with mass = 20 kg and radius = 0.10 m starts from...
A small, solid cylinder with mass = 20 kg and radius = 0.10 m starts from rest and rotates without friction about a fixed axis through its center of mass. A string is wrapped around the circumference of the cylinder and pulled using a constant force F. The resulting angular acceleration of the cylinder is 5.0 rad/s2. What's the angular velocity after 4.0 s, in radians per second? (The moment of inertia of the cylinder is 1 half M R...
A small, solid cylinder with mass = 20 kg and radius = 0.10 m starts from...
A small, solid cylinder with mass = 20 kg and radius = 0.10 m starts from rest and rotates without friction about a fixed axis through its center of mass. A string is wrapped around the circumference of the cylinder and pulled using a constant force F. The resulting angular acceleration of the cylinder is 5.0 rad/s2. (The moment of inertia of the cylinder is 1/2 MR^2.) 1. What's the force F, in Newtons? 2. What's the angular velocity after...
A solid cylinder of radius 0.5 m and mass 5.0 kg, initially at rest, starts to...
A solid cylinder of radius 0.5 m and mass 5.0 kg, initially at rest, starts to rotate about an axis through its center, with an angular acceleration of 0.2 rad/s^2. (a) Assuming a piece of string is wrapped around the cylinder, in such a way that the turning cylinder pulls the string onto itself, what is the total length of string wrapped on the cylinder at t= 10s? (b) What is the linear acceleration of a knot in the string...
A solid sphere of mass, M, and radius, R, is rigidly attached to a strong thin...
A solid sphere of mass, M, and radius, R, is rigidly attached to a strong thin rod of radius r that passes through the sphere at a distance of R/2. A string wrapped around the rod pulls with tension T. The rod's moment of inertia is negligible. (a) Find an expression for the sphere's angular acceleration. (b) If the sphere has a mass of 5.5 kg and a radius of 25 cm, how much tension must be applied to the...
A solid cylinder has a mass of 100 kg and radius 0.225m. The cylinder is attached...
A solid cylinder has a mass of 100 kg and radius 0.225m. The cylinder is attached to a frictionless horizontal axle. A long (light weight) cable is wrapped around the cylinder. Attached to the end of the cable is a 1.50 kg mass. The system is initially stationary. The hanging mass is then released. The mass pulls on the cable as it falls and this causes the cylinder to rotate. a) What is the velocity of the hanging mass after...
A solid uniform disk of mass M = 9.6 kg and radius R = 21 cm...
A solid uniform disk of mass M = 9.6 kg and radius R = 21 cm rests with its flat surface on a frictionless table (i.e., the axis of the cylinder is perpendicular to the table.) The diagram shows a top view. A string is wrapped around the rim of the disk and a constant force of F = 10.8 N is applied to the string. The string does not slip on the rim. 1) What is the acceleration of...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT