Exam 4 Review

12/02/2015

PHYSICS 2210 Fall 2015

Problem 1 (yf09-049) A thin, light wire is wrapped around the rim of a uniform disk of radius 𝑅=0.280 m, as shown. The disk rotates without friction about a stationary horizontal axis that passes through the center of the disk. An object of mass 𝑚=4.20 kg is suspended from the free end of the wire. The system is released from rest and the suspended object descends with constant acceleration. If the suspended object moves downward a distance of 3.00 m in 2.00 s, what is the mass 𝑀 of the disk? Solution (1/3): Consider forces acting on the mass 𝑚: Only forces in the y-direction (choose down as +y):

𝐹𝑦 = 𝑚𝑚 − 𝑇 = 𝑚𝑚 The disk is not free to move, but can rotate about the center:

𝐼𝛼 = 𝜏 Where 𝐼 = 𝑀𝑅2 2⁄ is the moment of inertia of the disk about its center, 𝛼 the angular acceleration. The only torque acting on the disk about its center is the tension force. For this problem we will choose clock-wise (CW) as the positive rotation sense

𝜏 = 𝑅𝑇 sin 90° = 𝑅𝑇 Note that the motion of m and the disk is linked by the string: and the relationship here (down and CW being positive in each case):

∆𝑦 = 𝑅∆𝜃 → 𝑣 = 𝑅𝜔 → 𝑚 = 𝑅𝛼 And so we will substitute 𝛼 = 𝑚 𝑅⁄ into 𝐼𝛼 = 𝜏

𝑚

𝑇

𝑅 𝑀

Problem 1 (yf09-049) A thin, light wire is wrapped around the rim of a uniform disk of radius 𝑅=0.280 m, as shown. The disk rotates without friction about a stationary horizontal axis that passes through the center of the disk. An object of mass 𝑚=4.20 kg is suspended from the free end of the wire. The system is released from rest and the suspended object descends with constant acceleration. If the suspended object moves downward a distance of 3.00 m in 2.00 s, what is the mass 𝑀 of the disk? Solution (2/3):

𝑀𝑅2

2 ∙𝑚𝑅

= 𝑅𝑇

Here we solve for 𝑇:

𝑇 =𝑚𝑅 ∙

𝑀𝑅2

2 ∙𝑚𝑅 =

12𝑀𝑚

Substitute this expression for T back into 𝐹𝑦 = 𝑚𝑚 − 𝑇 = 𝑚𝑚:

𝑚𝑚 = 𝑚𝑚 −12𝑀𝑚 → 2𝑚𝑚 = 2𝑚𝑚 −𝑀𝑚 → 𝑀𝑚 = 2𝑚𝑚 − 2𝑚𝑚

→ 𝑀 =2𝑚(𝑚 − 𝑚)

𝑚

Now we can determine the acceleration a from the given data: Starting at rest from 𝑦=0, we have 𝑦=3.00 m after t=2.00 s. Generally

𝑦 = 𝑦0 + 𝑣0𝑡 +12𝑚𝑡

2

𝑚

𝑇

𝑅 𝑀

Problem 1 (yf09-049) A thin, light wire is wrapped around the rim of a uniform disk of radius 𝑅=0.280 m, as shown. The disk rotates without friction about a stationary horizontal axis that passes through the center of the disk. An object of mass 𝑚=4.20 kg is suspended from the free end of the wire. The system is released from rest and the suspended object descends with constant acceleration. If the suspended object moves downward a distance of 3.00 m in 2.00 s, what is the mass 𝑀 of the disk? Solution (3/3): But we take 𝑦0=0, and we are given 𝑣0=0 (started from rest), and so

𝑦 =12 𝑚𝑡

2 → 𝑚 =2𝑦𝑡2 =

2(3.00 m)2.00 s 2 = 1.5 m s2⁄

We can now substitute the acceleration back into our equation for 𝑀

𝑀 =2𝑚(𝑚 − 𝑚)

𝑚 =2 4.20 kg 9.8 m s2⁄ − 1.5 m s2⁄

1.5 m s2⁄ = 46.5 kg

𝑚

𝑇

𝑅 𝑀

𝑚

𝑇

𝑀

𝜃

𝑥

ℓ

Problem 2 (cj18-037) The figure shows a box of mass 𝑀=50.0 kg hung from a horizontal beam. The beam has mass 𝑚=30.0 kg and length ℓ=1.40 m. The beam is attached to the wall on the left by a hinge. One the right, it is suspended by a rope from a second wall, at an angle 𝜃=143° as shown. The rope can withstand a maximum tension of 490 N. Find the maximum distance 𝑥 from the left that the box can be hung without breaking the rope. Solution (1/3): The system is in static equilibrium. We will choose the hinge at P to be the axis of rotation and we now enumerate the forces and the associated torques (assume CCW to be +): (1) Force by the hinge: 𝐹ℎ𝑥 in the x direction, 𝐹ℎ𝑦 in the y direction they exert no torque about P. (2) Weight of the beam: −𝑚𝑚 in the y direction only, and exerts torque at center of beam

𝜏𝑐 =ℓ2𝑚𝑚 sin −90° = −

ℓ2𝑚𝑚

(the vector from the hinge to the center-of-mass of beam points right, and the gravitational force points down, which is 90° in the CW (negative) direction (3) Weight of the box: −𝑀𝑚 in the y direction only, and exerts torque at right end of beam, and like the weight of the beam, it is also directed at − 90°. It acts at a distance x from the hinge.

𝜏𝐶 = 𝑥𝑀𝑚 sin −90° = −𝑥𝑀𝑚

P

𝑚

𝑇

𝑀

𝜃

𝑥

ℓ

Problem 2

Solution (2/3): (4) Tension force of the cable (whose magnitude we set to be the maximum value of 490 N). The force is directed at 𝜑 = 180° − 𝜃 = 180° − 143° = 37°, to the right and above the horizontal: the components are 𝑇 cos 37° in the x, and 𝑇 sin 37° in the y direction. The direction from the radial vector to the tension is also 𝜑 = +37° (at right end of beam)

𝜏𝑇 = ℓ𝑇 sin 37° Now we set the sum of force components in x and in y directions to zero:

𝐹ℎ𝑥 + 𝑇 cos 37° = 0 … 1 𝐹ℎ𝑥 − 𝑚𝑚 −𝑀𝑚 + 𝑇 sin 37° = 0 … 2

Setting sum of the torques to zero:

𝜏𝑐 + 𝜏𝐶 + 𝜏𝑇 = 0 → −ℓ2𝑚𝑚 − 𝑥𝑀𝑚 + ℓ𝑇 sin 37° = 0 … (3)

Solving for 𝑥 from equation (3) we get: −ℓ𝑚𝑚 − 2𝑥𝑀𝑚 + 2ℓ𝑇 sin 37° = 0 → 2𝑥𝑀𝑚 = 2ℓ𝑇 sin 37° − ℓ𝑚𝑚

P

(cj18-037) The figure shows a box of mass 𝑀=50.0 kg hung from a horizontal beam. The beam has mass 𝑚=30.0 kg and length ℓ=1.40 m. The beam is attached to the wall on the left by a hinge. One the right, it is suspended by a rope from a second wall, at an angle 𝜃=143° as shown. The rope can withstand a maximum tension of 490 N. Find the maximum distance 𝑥 from the left that the box can be hung without breaking the rope.

𝑚

𝑇

𝑀

𝜃

𝑥

ℓ

Solution (3/3): And so we have (everything is known on the right side:

𝑥 =2ℓ𝑇 sin 37° − ℓ𝑚𝑚

2𝑀𝑚 =2 1.40 m 490 N 0.60 − 1.40 m 30 kg 9.8 m 𝑠2⁄

2 50.0 kg 9.8 m 𝑠2⁄

𝑥 = 0.42 m

We actually did not need the force equations for this problem.

Problem 2 (cj18-037) The figure shows a box of mass 𝑀=50 kg hung from a horizontal beam. The beam has mass 𝑚=30 kg and length ℓ=1.4 m. The beam is attached to the wall on the left by a hinge. One the right, it is suspended by a rope from a second wall, at an angle 𝜃=143° as shown. The rope can withstand a maximum tension of 490 N. Find the maximum distance 𝑥 from the left that the box can be hung without breaking the rope.

P

Problem 3 (hr11-053) In the figure (overhead view), a uniform rod of length ℓ=0.500 m and mass 𝑀=4.00 kg can rotate in a horizontal plane about a vertical axis through its center. The rod is at rest when a 3.00 g bullet traveling in the plane is fired into one end of the rod. In the view from above, the bullet’s path makes angle 𝜃=60° with the rod (see figure). If the bullet lodges in the rod and the angular velocity of the rod is 10 rad/s immediately after the collision, what is the bullet’s speed 𝑣0 just before impact? Solution (1/3): • Consider the bullet + rod to be a system (this is actually necessary because the bullet

becomes lodged in the rod). • The collision is clearly not elastic (the bullet sticks to the rod). • The pivot at the rotation axis exerts an external force on the system, so linear momentum

is NOT conserved. • The force of the pivot acts at the rotation axis and exerts no torque, and so there is no net

external torque, and total angular momentum 𝐿 is conserved.

Before collision: the rod was at rest: no contribution to total 𝑣. The bullet has angular momentum given by 𝐿𝑏 = 𝑟 × �⃗�𝑏. In this case 𝑟 points along the rod from the rotation axis to the end struck by the bullet, i.e. to the right. The momentum �⃗�𝑏 of the bullet points up and to the right, so rotating from 𝑟 to �⃗�𝑏 rotates by angle 𝜑 = +𝜃 = +60°. And so 𝐿𝑏 points out of the page, or we say 𝐿𝑏 = +𝑟𝑝𝑏 sin 60° is positive (i.e. CCW)

𝜃 ℓ 2⁄

Problem 3 (hr11-053) In the figure (overhead view), a uniform rod of length ℓ=0.500 m and mass 𝑀=4.00 kg can rotate in a horizontal plane about a vertical axis through its center. The rod is at rest when a 3.00 g bullet traveling in the plane is fired into one end of the rod. In the view from above, the bullet’s path makes angle 𝜃=60° with the rod (see figure). If the bullet lodges in the rod and the angular velocity of the rod is 10 rad/s immediately after the collision, what is the bullet’s speed 𝑣0 just before impact? Solution (2/3): Alternate formula for angular momentum of bullet before collision: 𝐿𝑏 = +𝑚𝑣𝑚. It is positive because the bullet misses the axis to the right (traveling in the CCW sense around the axis). Here 𝑣 = 𝑣0, and 𝑚 is the minimum distance of approach, which is shown in the diagram added on this page. By trigonometry: 𝑚 = ℓ 2⁄ sin𝜃. And so the total angular momentum before the collision is given by:

𝐿𝑖 = 𝐿𝑏 = 𝑚𝑣0𝑚 = 𝑚𝑣0ℓ2 sin 60°

After the collision: The bullet and the rod become a system with combined moment-of-inertia given by

𝐼 = 𝐼𝑟𝑜𝑟 + 𝑚𝑟2 =1

12𝑀ℓ2 + 𝑚

ℓ2

2

=𝑀12 +

𝑚4 ℓ2 =

112 𝑀 + 3𝑚 ℓ2

And as the system rotates at 𝜔𝑓=10 rad/s after the collision, the final angular momentum is:

𝐿𝑓 = 𝐼𝜔𝑓 =1

12 𝑀 + 3𝑚 ℓ2𝜔𝑓

ℓ 2⁄

𝑚

Problem 3 (hr11-053) In the figure (overhead view), a uniform rod of length ℓ=0.500 m and mass 𝑀=4.00 kg can rotate in a horizontal plane about a vertical axis through its center. The rod is at rest when a 3.00 g bullet traveling in the plane is fired into one end of the rod. In the view from above, the bullet’s path makes angle 𝜃=60° with the rod (see figure). If the bullet lodges in the rod and the angular velocity of the rod is 10 rad/s immediately after the collision, what is the bullet’s speed 𝑣0 just before impact? Solution (3/3): Angular Momentum is conserved, so we set 𝐿𝑖 = 𝐿𝑓:

𝑚𝑣0ℓ2 sin 60° =

112 𝑀 + 3𝑚 ℓ2𝜔𝑓

Solving for 𝑣0 then gives

𝑣0 =2

𝑚ℓ sin 60° ∙1

12 𝑀 + 3𝑚 ℓ2𝜔𝑓 =𝑀 + 3𝑚 ℓ𝜔𝑓6𝑚 sin 60°

=4.00 kg + 3 ∙ 0.003 kg 0.500 m 10 rad s⁄

6 0.003 kg 3 2⁄= 1.28 × 103 m s⁄

Problem 4 (sj10-029) The figure shows a “cam”, which is a circular disk rotating on a shaft that does not pass through the center of the disk. For this cam, a uniform solid cylinder of radius R is first machined. Then an off-center hole of radius R/2 is drilled, parallel to the axis of the cylinder, and centered at a point a distance R/2 from the center of the cylinder. The cam, of mass M, is then slipped onto the circular shaft and welded into place. Find the kinetic energy of the cam when it is rotating with angular speed 𝜔 about the axis of the shaft?

Consider a piece that is a full cylinder consisting of the cam plus a piece that exactly plus the hole. The area of the hole is ¼ that of the full disk, and that of the cam is ¾ that the full disk. So the mass of the hole𝑀ℎ𝑜𝑜𝑜is related to the mass of the cam 𝑀 by

𝑀ℎ𝑜𝑜𝑜 = 𝑀 3⁄ And the mass of the full disk (FD) is therefore 𝑀𝐹𝐹 = 4𝑀 3⁄ . For this problem we will assume that the full disk is the sum of the cam and the “hole”, which also works for calculating the moment of inertia about the center of the hole (shaft):

𝐼𝐹𝐹 = 𝐼𝑐𝑐𝑐 + 𝐼ℎ𝑜𝑜𝑜

P. 1/3

Problem 4 (sj10-029) The figure shows a “cam”, which is a circular disk rotating on a shaft that does not pass through the center of the disk. For this cam, a uniform solid cylinder of radius R is first machined. Then an off-center hole of radius R/2 is drilled, parallel to the axis of the cylinder, and centered at a point a distance R/2 from the center of the cylinder. The cam, of mass M, is then slipped onto the circular shaft and welded into place. Find the kinetic energy of the cam when it is rotating with angular speed 𝜔 about the axis of the shaft?

But since the moment-of-inertia of a cylinder (both full disk and hole) is straight-forward, we will then calculate 𝐼𝑐𝑐𝑐 by

𝐼𝑐𝑐𝑐 = 𝐼𝐹𝐹 − 𝐼ℎ𝑜𝑜𝑜

We start with the hole: it is a uniform disk and rotates about its center:

𝐼ℎ𝑜𝑜𝑜 =12𝑀ℎ𝑜𝑜𝑜𝑅ℎ𝑜𝑜𝑜2 =

12 ∙

𝑀3 ∙

𝑅2

2=

124𝑀𝑅

2 𝑀ℎ𝑜𝑜𝑜 = 𝑀 3⁄

𝑅ℎ𝑜𝑜𝑜=𝑅 2⁄

P. 2/3

Problem 4 (sj10-029) The figure shows a “cam”, which is a circular disk rotating on a shaft that does not pass through the center of the disk. For this cam, a uniform solid cylinder of radius R is first machined. Then an off-center hole of radius R/2 is drilled, parallel to the axis of the cylinder, and centered at a point a distance R/2 from the center of the cylinder. The cam, of mass M, is then slipped onto the circular shaft and welded into place. Find the kinetic energy of the cam when it is rotating with angular speed 𝜔 about the axis of the shaft?

𝐼𝑐𝑐𝑐 = 𝐼𝐹𝐹 − 𝐼ℎ𝑜𝑜𝑜

Full Disk: by Parallel Axis Theorem: noting the actual axis of rotation is a distance 𝐷 = 𝑅/2 from the center of the full disk.

𝐼𝐹𝐹 = 𝐼𝐹𝐹(𝐶𝐶) + 𝑀𝐹𝐹𝐷2 =12𝑀𝐹𝐹𝑅2 + 𝑀𝐹𝐹

𝑅2

2

=124𝑀

3 𝑅2 +4𝑀

3𝑅2

4 =46 +

412 𝑀𝑅2 = 𝑀𝑅2

𝑀𝐹𝐹 = 4𝑀 3⁄

𝑅𝐹𝐹 = 𝑅

𝐷

𝐼𝑐𝑐𝑐 = 𝐼𝐹𝐹 − 𝐼ℎ𝑜𝑜𝑜 → 𝐼𝑐𝑐𝑐 = 𝑀𝑅2 −1

24𝑀𝑅2 =

2324𝑀𝑅

2

Rotating at angular velocity 𝜔

𝐾 =12 𝐼𝑐𝑐𝑐𝜔

2 =12

2324𝑀𝑅

2 𝜔2 = 2348𝑀𝑅

2𝜔2

P. 3/3