Chapter 9 - Rotational Dynamics Flashcards
A hiker, who weighs 985 N, is strolling through the woods and crosses a small horizontal bridge. The bridge is uniform, weighs 3610 N, and rests on two concrete supports, one at each end. He stops one-fifth of the way along the bridge. What is the magnitude of the force that a concrete support exerts on the bridge (a) at the near end and (b) at the far end?
Set the axis of rotation at one of the block so the F2 is 0 there, and the rotational motion occurs clockwise.
For net torque
T = -F1L + W(4/5L) + W(1/2L) = 0
You can eliminate the length of the bridge form every term, subsittuting in the known weights that gets F1 to come out to be 2590N
For net direction in the y
Fnety = F1-W-W+F2
You can know algebraically solve for F2 which is 2010N
Y
What holds true in static equilibrim?
Sum of clockwise torques = sum of counterclockwise torques
A uniform solid sphere of radius R and mass M = 5.0 kg rolls without slipping down a ramp (as shown in figure).
The ramp is inclined at an angle β = 30o to the horizontal. What is the numerical value of the sphere’s acceleration?
Translational motion (only consider Fx)
Fx = mgsinx - ffriction
ma = mgsinx - Ffriction
Rotational Motion (represented in two ways)
T = IA
and T = FfrictionR
Set them equal to each other
FfrictionR = IA
Sub in moment of inertia equation for sphere and equation for tangential acceleration
Ffriction R = 2/5Mr^2 * a/R
can cancel out the R’s and are left with
Ffriction = 2/5Ma
Now sub this into the F net x equation
mgsinx - 2/5Ma = ma
the mass is referencing the same thing so you can just cancel it out, solve and isolate for a, you should end up with
5/7gsinx = a
and a = 3.5
A cable is wrapped several times around a uniform solid cylinder
with mass M = 50 kg and radius R = 0.10 m and rotates freely without friction
about a horizontal axis. The bucket (mass m = 5 kg) must descend a height
h = 5 m to reach the water; it is suspended by a rope of negligible mass that
wraps around the cylinder. The winch (cylinder) handle falls off, releasing
the bucket to fall to the water, rotating the cylinder as it falls. Assuming that
the cable unwinds without stretching or slipping:
5 Find the magnitude of the angular velocity of the cylinder just before
the bucket hits the water.
5 Find the magnitude of the angular acceleration of the cylinder.
a) You’re going to use the conservation of mechnical energy, knowing that teh velocity initally of the cylinder is 0 which makes the inital kinetic energy and rotational kinetic energy 0
So you get
mgh = 1/2mv^2 + 1/2IW^2
you want to turn v into angular variables, so that would be wr (make sure to square both terms!!) and you awnt to turn I into moment of inertia for the shape which would be (1/2MR^2)
Now it’s easy subsitution and solving to get w = 40.4
**Below T tension and t torque
b) You’ve got F = ma
T - mg = ma
T = ma + mg
T = m(a +g)
You also got
t = TR
t = IA
Set them equal and sub in I
Sub in the equation for tension above
You get
m (RA + g) = 1/2MRA
Sub in your values and you should get a = 24.5
Cable is unwinding from winch. THe force is 9N, displacement is 2m, mass of winch is 50kg and radius is 0.06m. Find the angular velocity and the tangential velocity
What is important to recognize is that you’re given F and S which means you need to find work. And the work equals the change in kinetic energy. And you need to sub in the moment of interia for the shape given, to find w which is 20. And then you sub simply into the equation to get 1/2m/s for vtan
W = change in kinetic energy none initially so just final
1/2Iw^2 + 1/2mv^2
substitute v =rw and solve as so you should get
18=1/2w^2 (0.09 + 50 x 0.06^2)
w is 11.54
sub jnto v = rw to get v which is 0.69
Combined moment of inertia of a pully is 50kgm^2. Crate mass is 451kg, tension is 2150N and the lever arm of the string between motor and pulley is 0.6 and the lever arm of the string 2 between pulley and crate is 0.2m. Find the angular acceleration
1) You need to consider both the translational and rotational motion. So that would be the net force in the y direction and the net torque.
Fnety = T2 - mg = may
T2 = mg + may
Tnet = T1l1 - T2l2 = IA
(REALLY need to pay attention to the signs. So identify the direction of torque!!!!) The tension is pulling downwards so it pulls the pulley to the right hence the torque is negative
Sub in the T2 from the first equation to the second equation. For ay, it is l2a (radius is the lever arm)
You should get
T1L1 - (m(L2A+g) = IA
(1290 - 18.04A - 833.96) = 50A
A = 5.97
An 8.00-m ladder of weight WL=355 N leans against a smooth wall. The term
“smooth” means that the wall can exert only a normal force directed perpendicular
to the wall and cannot exert a frictional force parallel to it. A firefighter, whose
weight is WF = 875N, stands 6.30 m from the bottom of the ladder. The ladder makes a 50 degree angle with the ground. Assume that the
ladder’s weight acts at the ladder’s center and neglect the hose’s weight. Find the
forces that the wall and the ground exert on the ladder.
1) It would first be useful to draw a diagram labelling the 5 forces (x and y of the ground, weight of the ladder acting at the cneter and fireman, and the normal force. Set the axis of rotation around the ground so it cancels out these forces, and draw the lever arms from the other three forces towards the axis.
2) The system is in equilibrium because it is not moving. Hence set everything euqal to 0.
Fx = Gx - P = 0
Fy = Gy - Wl - Wf = 0
so Gy = 1230N
Tnet = -WlLl - WflF + PlP
you find the lever arms based on trigonometry of the triangles to find P which is teh unknown.
**you also want to make sure you define which direction torque is in and the lever arm applied will let you know which sign to put in your net equation
A string is wrapped around
a
uniform solid cylinder of mass M and radius R. The cylinder is
released from rest with the
string vertical and its top end
tied to
a fixed point, as shown
in Figure
. Find the magnitudes
of the acceleration and speed of
the center of gravity after the
cylinder has descended through
distance of
h= 0.25m
Basically some simple substiution, writing out the equations for rotational and translational motion.
Rotational:
t = IA
TR = (1/2MR^2)A
now can you subsitute rotational acceleration = a/R cancel out the R’s and you get
T = 1/2Ma
Translational:
T - Mg = -Ma
Sub in the value for T above, cancel out the m’s and isolate for a
mg-ma = 1/2Ma
you should get a = 6.53
To find the velocity, just use the simple kinematic equations, setting v = 0, you know the acceleration, y = 0.25 and you slayyyy
a 5.0 kg dumbbell in each hand. He
is set rotating about a vertical axis, making one revolution in 2.0 s.
His moment of inertia (without the dumbbells) is 3.0 kg.m2 when his
arms are outstretched and drops to 2.2 kg.m2 when his arms are pulled in
close to his chest. The dumbbells are 1.0 m from the axis initially and
0.20 m from it at the end. Find the professor’s new angular velocity if he
pulls the dumbbells close to his chest and compare the final total kinetic
energy with the initial value.
This is a conservation of momentum problem, identifiable by the fact that the moment of inertia changes, and its focusing on just the rotational motion, there is no translational motion.
1) First calculate the inital moment of inertia and final moment of inertia. You have to factor in both that of the professor and the dumbells since they both make up the system. The equation is mr^2, and since there are two dumbells it will be 2mr^2. You should get 13 and 2.6 respectively
2) The inital angular velocity can be found using the period, and it is pi.
3) Then you sub into Iwf = Iwi and then isolate and solve for the wf
4) Then just sub this value into the kinetic energy equations (just considering rotational motion to find the kinetic energy before and after
When determinign where the lever arm is, how do you do tihs?
It is always perpendicular to the force applied and connects to teh axis of rotation
Equation for center of graivty
Xcg = (W1X1 + W2X2)/W1 + W2
What is the equation for rotational work?
Wr = Tx
where x is the angular displacement
When is angular momentum conserved?
When the net torque is 0
