We know that a net force applied to a body gives that body an acceleration. But what does it take to give a body an angular acceleration?
Ch 10 – learning goals an acceleration. But what does it take to give a body an
The quantitative measure of the tendency of a an acceleration. But what does it take to give a body an force to change a body’s rotational motion is called torque;
Fa applies a torque about point O to the wrench.
Fb applies a greater torque about O,
and Fc applies zero torque about O.
τ = Fl
l: the perpendicular distancel1 between point O and the line of action of the force
The direction of torque an acceleration. But what does it take to give a body an
If an acceleration. But what does it take to give a body an φ is angle between force F and distance r
τ = F∙(r∙sin)
r∙sin - perpendicular distance
τ = r∙(F∙sin)
F∙sin - perpendicular force
Magnitude an acceleration. But what does it take to give a body an :
When a force F acts at a point having a position vector r with respect to an origin O, the torque of the force with respect to O is the vector quantity
The direction of torque is perpendicular to both r and F. The torque vector is directed along the axisof rotation, with a sense given by the right-hand rule.
A dot ● means pointing an acceleration. But what does it take to give a body an out of the screen
A cross × means pointing into the screen
45 an acceleration. But what does it take to give a body an o
A, C, B
since an acceleration. But what does it take to give a body an
We write an equation like this for every particle in the body and then add all these equations:
Just as Newton’s second law says that the net force on a particle equals the particle’s mass times its acceleration, the equation says that the net torque on a rigid body equals the body’s moment of inertia about the rotation axis times its angular acceleration.
Consider the situation on the diagram, find the acceleration of the block of mass m.
The point on the wheel that contacts the surface must be instantaneously at rest so that it does not slip. Hence the velocity v1’ of the point of contact relative to the center of mass must have the same magnitude but opposite direction as the center-of-mass velocity vcm. If the radius of the wheel is R and its angular speed about the center of mass is ω, then the magnitude of v1’ is R∙ω; hence we must have
vcm = Rω (condition for rolling without slipping)
When a rigid body with total mass M moves, its motion can be described by combining translational motion and rotational motion
In translation, the acceleration acm of the center of mass is the same as that of a point mass M acted on by all the external forces on the actual body:
The rotational motion about the center of mass is described by the rotational analog of Newton’s 2nd law:
When a perfectly rigid sphere is rolling down a perfectly rigid incline, there is no sliding at the point of contact, so friction does no work. However, in reality, when a not so perfectly rigid sphere rolling down a not so perfectly rigid incline, there are some deformations at the points of contact. As a result, there is rolling friction.
Often the rolling body and the surface are rigid enough that rolling friction can be ignored.
The work is replaced by a hollow cylinder of the same mass and radiusdW done by the force Ftan while a point on the rim moves a distance ds is dW = Ftan∙ds. If dθ is measured in radians, then ds = R∙dθ
The work done by a constant torque is the product of torque and the angular displacement. If torque is expressed in Newton-meters and angular displacement in radian, the work is in joules.
Only the tangent component of force does work, other components do no work.
When a torque does work on a rotating rigid body, the kinetic energy changes by an amount equal to the work done.
Wtot = ½ I∙ω22 – ½ I∙ω12
P = dW/dt = τz(dθ/dt) = τzω
When a torque acts on a body the rotates with angular velocity ωz, its power is the product of τz and ωz. This is the analog of the relationship P = F∙v
∑F = ma which has a moment of inertial twice as large as the other cylinder. Each cylinder is initially at rest after one complete rotation, which cylinder has the greater kinetic energy?
K = ½ mv2
Wtot = ½ mv22 - ½ mv12
P = Fv
L = I ∙ ω
The value of L depends on the choice of origin O, since it involves the particle’s position vector relative to O. the units of angular momentum are kg∙m2/s.
If at moment t the position of the particle is r and its linear momentum is p = m∙v, then the magnitude of angular momentum is:
L = (mvsinΦ)r
L = mv(rsinΦ) = mvl
Where l is the perpendicular distance from the line of v to O. this distance plays the role of “lever arm” for the momentum vector.
= F the angular momentum vector net
The derivation of the angular momentum vector
If we take the time derivative of , using the rule for the derivative of a product:
Consider a rigid body rotating about the z-axis with angular speed ω. First let’s think about a thin slice of the body lying in the xy-plane.
Each particle moves in a circle centered at the origin, and at each instant its velocity vi is perpendicular to its position vector ri. Hence Φ = 90o for every particle. A particle with mass mi at a distance ri from O has a speed vi = riω. The magnitude of its angular momentum is:
Li = (mivisinΦ)ri = (miriω)ri = miri2ω
The direction of each particle's angular momentum, as given by the right-hand rule for the vector product, is along the +z-axis.
We can do this same calculation for the other slides of the body, all parallel to the xy-plane. Since the rigid body is rotating the axis of symmetry, their angular momentum vector sum L1 + L2 also lies along the symmetry, and its magnitude is L = Iω.
Since ω and L have the same direction, we have a vector relationship: L = Iω
Where I is the moment of inertia of the slice about the z-axis.
If the system of particles is a rigid body rotating about a symmetry axis (z-axis), the Lz = Iωz and I is constant. If this axis has a fixed direction in space, then the vectors L and ω change only in magnitude, not in direction. dLz/dt = Idωz/dt = Iαz ∑τ = Iαz
When the net external torque acting on a system is zero, the total angular momentum of the system is constant (conserved).
I the other end of the string and whirl the ball in a circle around your hand. 1ω1z = I2ω2z
When a system has several parts, the internal forces that the parts exert on each other cause changes in the angular momenta of the parts, but the total angular momentum doesn’t change. The total angular momentum of the system is constant.
(hint: use angular momentum ideas. Assume that the sun, moon, and planets exert negligibly small torques on the earth.)
L = Iω
I increases, ω decreases, days is longer.
example warming, the melted ice would redistribute itself over the earth. This change would cause the length of the day (the time needed for the earth to rotate once in its axis) to
r warming, the melted ice would redistribute itself over the earth. This change would cause the length of the day (the time needed for the earth to rotate once in its axis) to