Abeka Physics Section 13.3 pg. 195 - 199 Flashcards
Rate of angular acceleration when a body is set into rotation by a force depends on what
Rate of angular acceleration when a body is set into rotation by a force depends on where the force is applied.
Define and describe radius of a force
> also called its torque arm
is vector directed from center of rotation to point where force is applied, with magnitude equal to distance between these points
When does Greek lowercase “tau” (torque) = Fr?
tau = Fr when the force is perpendicular to the radius
What are the units of torque
units of torque
>N x m (SI): cannot be simplified to joules because radius is not parallel to applied force
>lb x ft (US customary)
With what does rotary motion arise?
> rotary motion cannot arise without action of one or more torques
like when the volume knob on the face of car radio is twisted by your fingers because each finger applies a torque to rim of the knob
never does rotation occur if force is applied directly to rotational axis
instead, force must be applied at a distance from axis so that the radius of the force is greater than zero; in other words, force must produce a torque
Describe the law of torque
> when Newton’s second law of motion (F=ma) is restated for rotary motion, the law becomes tau = I x alpha
where tau is analogous to F, I is analogous to m, and alpha is analogous to a
this equation is called law of torque
equation shows that for body of given rotational inertia, changes in torque bring proportional changes in angular acceleration
equation also shows that torque necessary to produce given rate of angular acceleration is determined by rotational inertia of body
Solve this: What torque is required to accelerate a wheel from rest to 100.rev/s in 30.0s if the mass of the wheel is 10.0kg? Assume that the wheel has an inner radius, r(sub 1), equal to 0.950m and an outer radius, r(sub 2), equal to 1.00m.
Use equation tau=I x alpha, which requires the rotational inertia and angular acceleration of the wheels. From Figure 13.9, the solid with the shape most similar to the wheel is the disk with a central hole. Substitute the given values of m (10.0kg), r(sub 1) (0.950m), and r(sub 2) (1.00m).
I = 1/2 m[r(sub 1)^2 + r(sub 2)^2]
I = 1/2 (10.0kg)[(0.950m)^2 + (1.00m)^2]
I = 9.5125 kg x m^2 = 9.51 kg x m^2
Find the angular acceleration of the wheel by using equation alpha(sub av) = omega(sub f) - omega(sub i) / t, where omega(sub i) is 0rev/s, omega(sub f) is 100.rev/s, and t is 30.0s.
alpha(sub av) = 100.rev/s - 0rev/s / 30.0s
alpha(sub av) = 3.333… rev/s^2
then, convert to rad/s^2
alpha = (3.333…rev/s)(2pi rad/1 rev)
alpha = 20.943… rad/s^2
Finally, substitute into equation tau = I x alpha
tau = (9.5125 kg x m^2)(20.943…rad/s^2)
tau = 199 N x m
Define torque
> torque also referred to as moment of force
torque - quantitative measurement of the effect of whenever force and its radius are perpendicular and the applied force is rotational
torque is calculated by multiplying radius by perpendicular force
tau = Fr
Give the equations for work in linear motion and work in angular motion
> work in linear motion: W = Fd
>work in angular motion: W = “tau” x theta(sub rad)
Give equations for power in linear motion and in angular motion
> power in linear motion: P = Fd/t = Fv
power in angular motion:
P = Fr x theta(sub rad)/t = “tau” x omega
Give equations for kinetic energy in linear motion and angular motion
> kinetic energy in linear motion: KE = 1/2 x mv^2
>kinetic energy in angular motion: KE = 1/2 x l x omega^2
Give equations for momentum in linear motion and in angular motion
> momentum in linear motion: p = mv
>momentum in angular motion: L = l x omega
Describe angular momentum
> also known as rotary momentum
symbolized as L
L = l x omega
Describe total kinetic energy of a wheel on a moving bicycle
> wheel is simultaneously spinning and moving forward
if forward translational motion is uniform (at constant speed), rotary motion must be uniform also
total kinetic energy of the wheel is the sum of its rotational and translational kinetic energies:
KE = 1/2 mv^2 + 1/2 I x omega^2
Describe conservation of energy in rotary motion
> total potential and kinetic energy of any object is conserved unless it is subjected to nonconservative forces like friction
example: if wheel rolls down frictionless slope, gain in kinetic energy equals loss in gravitational potential energy
since PE = mgh and g is constant, loss in PE depends solely on the mass of of the wheel and on the distance of the vertical drop
thus, any two wheels of equal mass must acquire the same KE when they roll down the same slope
