In Fig. 11-60, a constant horizontal force of magnitude 12 N is applied to a uniform solid cylinder by fishing line wrapped around the cylinder. The mass of the cylinder is 10kg, its radius is 0.10 m, and the cylinder rolls smoothly on the horizontal surface. (a) What is the magnitude of the acceleration of the center of mass of the cylinder? (b) What is the magnitude of the angular acceleration of the cylinder about the center of mass? (c) In unit-vector notation, what is the frictional force acting on the cylinder?
Use formula for torque to find angular acceleration. Use angular acceleration to find linear acceleration, and finally use Newton’s second law to find friction force.
Formula are as follow:
Where, is torque, is force, is moment of inertia, is radius, is angular acceleration and is acceleration.
To calculate acceleration of center of mass, first calculate angular acceleration as follows:
Moment of inertia about point of contact is given by parallel axis theorem as follows:
Force is applied at the top of the cylinder, considering the bottom of the cylinder as pivot point and r as the radius of the cylinder,
Now, acceleration of center of mass is as follows:
Hence, acceleration of center of mass is .
From above calculations,
Hence, angular acceleration of center of mass is .
Frictional force (F) in unit vector form is as follows:
Now, to find friction force, use Newton’s second law of motion,
So, friction force is as follows:
In unit vector notation,
Since, value F is positive, it is along the positive x axis.
Hence, frictional force in terms of unit vector notation is .
Therefore, the formula for torque to find the angular acceleration can be used. Using angular acceleration, linear acceleration can be found. Newton’s second law can be used to find friction force.
A particle is acted on by two torques about the origin: has a magnitude of and is directed in the positive direction of the axis, and has a magnitude of and is directed in the negative direction of the y axis. In unit-vector notation, find , where is the angular momentum of the particle about the origin.
Question: In Figure, a solid cylinder of radius 10 and mass 12 kg starts from rest and rolls without slipping a distance L =6.0 m down a roof that is inclined at the angle . (a) What is the angular speed of the cylinder about its centre as it leaves the roof? (b) The roof’s edge is at height H = 5.0m How far horizontally from the roof’s edge does the cylinder hit the level ground?
Figure shows a rigid structure consisting of a circular hoop of radius and mass , and a square made of four thin bars, each of length and mass . The rigid structure rotates at a constant speed about a vertical axis, with a period of rotation of . Assuming and role="math" localid="1660971946053" ,
(a) Calculate the structure’s rotational inertia about the axis of rotation?
(b) Calculate its angular momentum about that axis?
Figure shows three rotating, uniform disks that are coupled by belts. One belt runs around the rims of disks A and C. Another belt runs around a central hub on disk A and the rim of disk B. The belts move smoothly without slippage on the rims and hub. Disk A has radius R; its hub has radius ; disk B has radius ; and disk C has radius Disks B and C have the same density (mass per unit volume) and thickness. What is the ratio of the magnitude of the angular momentum of disk C to that of disk B?
Figure 11-24 shows two particles A and B at xyz coordinates (1 m, 1 m, 0) and (1 m, 0, 1 m). Acting on each particle are three numbered forces, all of the same magnitude and each directed parallel to an axis. (a) Which of the forces produce a torque about the origin that is directed parallel to y? (b) Rank the forces according to the magnitudes of the torques they produce on the particles about the origin, greatest first.
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