Example 16.1 (3/4) A round βwheelβ of mass π, radius π and π°πͺπͺ = πΈπͺπΉπ (0 < πΎ < 1) rolls without slipping down an incline at angle π from the horizontal. Find the linear acceleration of the wheel down the incline.
Alternate Solution: Let +x be down slope and +y point perpendicularly out of the incline. We will chose CCW As the positive rotation sense. For rolling without slipping then we have:
π£π₯ = βπ π, ππ₯ = βπ π Whereπ£π₯ and ππ₯ are the velocity and acceleration components down slope. Applying Newtonβs Second Law on the CM, first in the y-direction
πππ¦ = π βππ cosπ = 0 β π = ππ cosπ Which is not really used in this problem except possibly to verify that ππ < ππ π In the x-direction :
πππ₯ = ππ sinπ β ππ β¦ 1 Next we look at rotation about the CM:
πΌπΆπΆπ = π ππ sin(β90Β°) = βπ ππ *** We have used the fact that, for the purpose of calculating torque, the force of gravity acts at the center-of-mass (CM). This follows from πΌπΆπΆπ = π ππ sin(β90Β°) = βπ ππ
π
ππ
π ππ
π πΌπΆπΆ = πΎππ 2
π₯
π¦
Example 16.1 (4/4) A round βwheelβ of mass π, radius π and π°πͺπͺ = πΈπͺπΉπ (0 < πΎ < 1) rolls without slipping down an incline at angle π from the horizontal. Find the linear acceleration of the wheel down the incline.
Solution (continued): From last page πππ₯ = ππ sinπ β ππ β¦ 1 πΌπΆπΆπ = π ππ sin(β90Β°) = βπ ππ
But ππ₯ = βπ π ,β π = βππ₯ π β , and so we have
πΎππ 2 ββππ₯π = π ππ , β βπΎπππ₯ = ππ β¦ (2)
substituting (2) back into (1) for ππ : πππ₯ = ππ sinπ β πΎπππ₯, β 1 + πΎ πππ₯ = ππ sinπ
And so we have:
ππ₯ =π sinπ 1 + πΎ
Example given in Main Point: Solid Ball/Sphere
πΌ =25ππ 2 β πΎ =
25
β 1 + πΎ =75
β 1
1 + πΎ=
75
ππ₯ =75π sinπ
Example 16.2 (1/6) A 1152 kg car is being unloaded by a winch. At the moment shown below, the gearbox shaft of the winch breaks, and the car falls from rest. During the car's fall, there is no slipping between the (massless) rope, the pulley, and the winch drum. The moment of inertia of the winch drum is 344 kgΒ·m2 and that of the pulley is 3 kgΒ·m2. The radius of the winch drum is 0.80 m and that of the pulley is 0.30 m. Find the speed of the car as it hits the water.
Solution: Looking for the speed of something as it drops through a certain distance. Does it remind you of a problem to solve using conservation of energy? Except in this case, as the car acquires speed, both the pulley and the drum acquire angular speed (no slipping!!!): Drum: π π·ππ· = π£πΆ ... (1), Pulley: π πππ = π£πΆ β¦ (2). The total kinetic energy of the system is the sum of those of the three objects:
πΎ = πΎπ· + πΎπ + πΎπΆ =12 πΌπ·ππ·
2 +12 πΌπππ
2 +12ππΆπ£πΆ2 β¦ (3)
Substituting (1) and (2) into (3) gives us
πΎ =12 πΌπ·
π£πΆπ π·
2
+12 πΌπ
π£πΆπ π
2
+12ππΆπ£πΆ2 =
12
πΌπ·π π·2
+πΌππ π2
+ ππΆ π£πΆ2
Example 16.2 (2/6) A 1152 kg car falls: no slipping between the (massless) rope, the pulley (has mass and rotates with the string), and the winch drum. Moment of inertia of the winch drum: 344 kgΒ·m2, of pulley: 3 kgΒ·m2. Radius of the winch drum is 0.80 m, pulley: 0.30 m. Find the speed of the car as it hits the water.
πΎ =12
πΌπ·π π·2
+πΌππ π2
+ ππΆ π£πΆ2
No friction: total energy is conserved. The only potential energy is the gravitational potential energy of the car: so we have βπΈ = βπΎ + βπ = 0. We started from rest: πΎπ = 0 So the final kinetic energy (after 5.0m drop) is then
πΎπ = πΎπ + βπΎ = ββπ = β ππΆπββ , ββ = β5.0m And so
12
πΌπ·π π·2
+πΌππ π2
+ ππΆ π£πΆπ2 = β ππΆπββ
π£πΆπ2 =β2ππΆπββ
πΌπ·π π·2
+ πΌππ π2
+ ππΆ
=β2 β 1152kg β 9.8 m s2β β (β5.0m)344 kgΒ·m2
0.80 m 2 + 3 kgΒ·m2
0.30 m 2 + 1152kg=
112896 kg β m2 s2β1723.8 kg
= 65.53m2
s2
π£πΆπ = 8.10 m sβ
Unit 17
Remember from Last class, we said that the gravitational potential energy ππof an object is given by the height of the Center-of-mass. If we take ππ = 0 at π¦ = 0 (where the +y direction is up), then
ππ = ππππΆπΆ The Main Point above is equivalent to this statement about ππ
Torque of Gravitational Force Let the pivot for the torque calculation be at origin (this is not required). We think of an extended object as composed of point masses π1,π2,β― ,ππ located at π1, π2 β― , ππ. Then the torque on the body is the sum of the torque on the point masses (remember this is again a vector equation: it is the shorthand for 3 equations)
π = οΏ½ πππ
π=1= οΏ½ (ππ Γ οΏ½βοΏ½π)
π
π=1
For gravity, then οΏ½βοΏ½π=πποΏ½βοΏ½, where οΏ½βοΏ½ is the gravitational acceleration vector (usually written as οΏ½βοΏ½ = βποΏ½ΜοΏ½, if we take +π¦ to be up)
π = οΏ½ (ππ Γ πποΏ½βοΏ½π
π=1) = οΏ½ (ππππ Γ οΏ½βοΏ½
π
π=1) = οΏ½ ππππ
π
π=1Γ οΏ½βοΏ½
=1ποΏ½ ππππ
π
π=1Γ ποΏ½βοΏ½ = π πΆπΆ Γ οΏ½βοΏ½π
Where we have used the fact that cross products are distributive
Poll 11-16-01
In case 1, one end of a horizontal massless rod of length L is attached to a vertical wall by a hinge, and the other end holds a ball of mass M. In case 2 the massless rod is twice as long and makes an angle of 30Β° with the wall as shown.
In which case is the total torque about an axis through the hinge biggest?
A. Case 1 B. Case 2 C. Same
Unit 17
These are 2D problems for Force (x and y-): βπππ = π, βπππ = π, and 1D problems for torque (CCW+, CWβ): βππ = π And gives you a total of THREE (3) equations.
Treat the x- and y-components of the hinge force separately
πΉπ»π₯
πΉπ»π¦
Poll 11-16-02
An object is made by hanging a ball of mass M from one end of a plank having the same mass and length L. The object is then pivoted at a point a distance L/4 from the end of the plank supporting the ball, as shown below.
Is the object balanced?
A. Yes B. No, it will fall to the left. C. No, it will fall to the right.
πΉπ = ππ
Poll 11-16-03
In case 1, one end of a horizontal plank of mass M and length L is attached to a wall by a hinge and the other end is held up by a wire attached to the wall. In case 2 the plank is half the length but has the same mass as in case 1, and the wire makes the same angle with the plank.
In which case is the tension in the wire biggest?
A. Case 1 B. Case 2 C. Same
Unit 17
This is exactly the same as π = π Γ π Some people have found the βlever armβ confusing A different way of looking at it: The line-of-action of the force is a line parallel to the direction of the force and passing through the point of application of the force) The βLever armβ, πβ₯, is the perpendicular distance (of closest approach) from the rotation axis to the line-of-action
line-of-action πβ₯
Lever Arm
Example 17-1 A traffic light hangs from a beam as shown in the figure. The uniform aluminum beam AB is 7.20 m long and has a mass of 12.0 kg. The mass of the traffic light is 21.5 kg. Determine (a) the tension in the horizontal massless cable CD, and (b) the vertical and horizontal components of the force exerted by the pivot A on the aluminum beam.
Solution This problem is all about balancing all forces and torques on the beam AB. We solve such problems by making a catalogue of all forces acting on the beam, and the torque each exerts. But first we must decide on a rotation axis about which to calculate torque. β’ It is usually convenient to pick a βpivotβ that exerts force in both x and y directions on the
element in question: because then these forces exert NO torque!!! We pick A (1) Force of pivot: πΉππ₯ in the x-direction, πΉππ¦ in the y direction. Notice we are treating them
like two separate forces for convenience. They exert no torque (they act at the chosen rotation axis).
(2) Tension Force in the cable: π directed in the βx direction, The line-of-action is horizontal, The lever arm, πβ₯, is given in the diagram to be 3.80m. The torque (it acts CCW so it is POSITIVE) is ππ = +πβ₯π = + 3.80π π
π line-of-action
πΉππ₯
πΉππ¦
πβ₯= Lever Arm
Example 17-1
(3) Weight of the beam itself. ππ in the βy direction Torque: ππ acts at radius π=β 2β =3.60 m. Angle from radial vector to weight is β127Β° So the torque is π π = β 2β ππsin β127Β° : Note the negative orientation of this torque (it wants to deflect beam CW around the pivot) is contained in sin β127Β° = β sin 127Β° (4) Weight of traffic light: Mπ in the βy direction Acts at distance π=β=7.20 m, and angle from radial vector to weight is AGAIN β127Β° π πΆ = βππ sin β127Β° Force components in the x- and y-direction independently sum to zero:
πΉπ₯ = πΉππ₯ β π = 0 β¦ (1) πΉπ¦ = πΉππ¦ β ππ βππ = 0 β¦ 2
(*** note you CANNOT add the magnitudes of the forces) Torques add to zero
π = ππ + π π + π πΆ = 3.80 m π β 3.60 m ππ sin 127Β° β 7.20 m ππ sin 127Β° = 0 β¦ (3)
πΉππ₯
πΉππ¦
π
ππ
37Β° 53Β° 127Β°
ππ Uniform beam AB: β=7.20 m, π =12.0 kg. (gi09-019) Traffic light π=21.5 kg. Determine (a) tension π (b) components πΉππ₯, πΉππ¦ of the force exerted by the pivot on the beam
Example 17-1 Uniform beam AB: β=7.20 m, π =12.0 kg. (gi09-019) Traffic light π=21.5 kg. Determine (a) tension π (b) components πΉππ₯, πΉππ¦ of the force exerted by the pivot on the beam
From equation (3) we have
π =3.60 m ππ sin 127Β° + 7.20 m ππ sin 127Β°
3.80 m = π sin 127Β°3.60 m π + 7.20 m π
3.80 m= 9.8 m s2β 0.799
3.60 m 12.0 kg + (7.20 m)(21.5 kg)3.80 m
= 7.827 m s2β198 kg β m
3.80 m = 408 N
From equation (1): πΉππ₯ β π = 0 β πΉππ₯ = π = 408 N
From equation (2): πΉππ¦ β ππ βππ = 0
πΉππ¦ = ππ + ππ = 12.0 kg + 21.5 kg 9.8 m s2β = 328 N
Unit 18
We already covered this in Unit 16
This statement is somewhat misleading: because πππ πππΆπΆβ = ππ is never ZERO JUST IGNORE IT!!!!
π
What we really want to say is πππ ππβ = 0
To minimize the potential energy
ππ = ππππΆπΆ
= ππβ2
sinπ + π
π
πΉππ₯
πΉππ¦
π
ππ
37Β° 53Β° 127Β°
ππ
Example 17-1 synopsis A traffic light hangs from a beam as shown in the figure. The uniform aluminum beam AB is 7.20 m long and has a mass of 12.0 kg. The mass of the traffic light is 21.5 kg. Determine (a) the tension in the horizontal massless cable CD, and (b) the vertical and horizontal components of the force exerted by the pivot A on the aluminum beam.
Choose pivot at A to be rotation axis. (1) Force components πΉππ₯ in the +x direction, πΉππ¦ in the +y direction. They exert no torque (2) Tension Force : π directed in the βx direction ππ = +π β (3.80m) (3) Weight of the beam: ππ in the βy direction , torque is π π = β β 2β ππsin 127Β° : (4) Weight of traffic light: Mπ in the βy direction ππΆ = ββππ sin 127Β°
πΉπ₯ = πΉππ₯ β π = 0 β¦ (1) πΉπ¦ = πΉππ¦ β ππ βππ = 0 β¦ 2
π = ππ + π π + π πΆ = 3.80 m π β 3.60 m ππ sin 127Β° β 7.20 m ππ sin 127Β° = 0 β¦ 3
π = 9.8 m s2β 0.7993.60 m 12.0 kg + (7.20 m)(21.5 kg)
3.80 m = 408 N
πΉππ₯ β π = 0 β πΉππ₯ = π = 408 N
πΉππ¦ β ππ βππ = 0 πΉππ¦ = ππ + ππ = 12.0 kg + 21.5 kg 9.8 m s2β = 328 N
Example 18-1 A uniform box of mass π=90 kg, height π»=1.2 m and with π=0.80 m sits on a rough, horizontal floor. You can apply a horizontal force πΉπ΄ at a single point on the left side of the box (say, at distance β above the floor). The static coefficient of friction between the box and the floor is ππ =1.10. (a) Find the minimum force πΉπ΄ needed to move the box. (b) Find the maximum distance π above the floor at which you can apply this minimum force without tipping the box over.
We start with a free-body diagram of the box. The additional forces are (1) The weight of the box ππ (2) The normal force π (3) The static friction force We start by balancing the forces, assuming the system is just on the verge of slipping, so that we still have static equilibrium:
πΉπ¦ = π βππ = 0, β π = ππ πΉπ₯ = πΉπ΄ β ππ = 0, β πΉπ΄ = ππ
We are on the verge of slipping ππ = ππ π = ππ ππ, and
πΉπ΄ = ππ ππ = 1.10 90 kg 9.8 m s2β = 970 N
π»
π
β
πΉπ΄ π
ππ
π»
π
β
πΉπ΄ π
ππ
ππ π
ππ
Example 18-1 π=90 kg, π»=1.2 m, π=0.80 m, ππ =1.10. (a) Find the minimum force πΉπ΄ needed to move the box. (b) Find the maximum β above the floor at which you can apply this minimum force without tipping the box over.
We now examine the torques exerted by the forces about the LOWER RIGHT CORNER (it is the pivot for tipping). (1) Applied force: ππ΄ = ββπΉπ΄ We have used a trick here: in finding the torque π = ππΉ sinπ, we can also try to identify the distance π sinπ, which is often easy to quantify. Here we have π sin 180Β° β π = π sinπ = β (2) Weight: ππΆ = + π 2β ππ Here we have π sin 180Β° β π = π sinπ = π 2β Where do the normal force and the friction forces act? The effecitve point of action lies somewhere on the bottom surface of the box. We think of this as the βweightβ of box βshiftingβ as we increase the force πΉπ΄
π
β
πΉπ΄ π
ππ π π sinπ
= β
π
180Β° βπ
π
πΉπ΄ π
ππ π
π sinπ = π 2β
π
180Β° βπ
Example 18-1
If πΉπ΄ = 0 then the normal forces (and hence the friction) acts at the center of the bottom: the torques from the weight and normal force cancel exactly. As we increase πΉπ΄, the point of action of the normal force (and hence the friction) shifts to the right. When on the verge of tipping, they act at the lower right corner: hence they exert no torque for part (b) Setting the net torque about lower-right corner to zero:
π = ππ΄ + ππΆ = ββπΉπ΄ + π 2β ππ = 0
β =π 2β ππ
πΉπ΄=
0.40 m (90 kg) 9.8 m s2β 970 N
= 0.364 m Answer (a) minimum force of 970 N (b) Maximum height (with 970 N) of 0.364 m
π»
π
π
ππ
ππ
π
π=90 kg, π»=1.2 m, π=0.80 m, ππ =1.10. (a) Find the minimum force πΉπ΄ needed to move the box. (b) Find the maximum β above the floor at which you can apply this minimum force without tipping the box over.
π»
π
β
πΉπ΄ π
ππ
π
ππ
ππ