PHYSICS 2210 Fall Exam 4 Review 12/02/2015

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Loreen McDowell
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1 PHYSICS 10 Fall 015 Exa 4 Review 1/0/015

2 (yf09-049) A thin, light wire is wrapped around the ri of a unifor disk of radius R=0.80, as shown. The disk rotates without friction about a stationary horizontal axis that passes through the center of the disk. An object of ass =4.0 kg is suspended fro the free end of the wire. The syste is released fro rest and the suspended object descends with constant acceleration. If the suspended object oves downward a distance of 3.00 in.00 s, what is the ass M of the disk? Solution (1/3): Proble 1 Consider forces acting on the ass : Only forces in the y-direction (choose down as +y): F y = T = The disk is not free to ove, but can rotate about the center: Iα = τ Where I = MR is the oent 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 proble we will choose clock-wise (CW) as the positive rotation sense τ = RT sin 90 = RT Note that the otion of 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 = Rα And so we will substitute α = R into Iα = τ M R T

3 (yf09-049) A thin, light wire is wrapped around the ri of a unifor disk of radius R=0.80, as shown. The disk rotates without friction about a stationary horizontal axis that passes through the center of the disk. An object of ass =4.0 kg is suspended fro the free end of the wire. The syste is released fro rest and the suspended object descends with constant acceleration. If the suspended object oves downward a distance of 3.00 in.00 s, what is the ass M of the disk? Solution (/3): Here we solve for T: MR R = RT T = R MR R = 1 M Substitute this expression for T back into F y = T = : Proble 1 M R T = 1 M = M M = ( ) M = Now we can deterine the acceleration a fro the given data: Starting at rest fro y=0, we have y=3.00 after t=.00 s. Generally y = y 0 + v 0 t + 1 t

4 (yf09-049) A thin, light wire is wrapped around the ri of a unifor disk of radius R=0.80, as shown. The disk rotates without friction about a stationary horizontal axis that passes through the center of the disk. An object of ass =4.0 kg is suspended fro the free end of the wire. The syste is released fro rest and the suspended object descends with constant acceleration. If the suspended object oves downward a distance of 3.00 in.00 s, what is the ass M of the disk? Solution (3/3): But we take y 0 =0, and we are given v 0 =0 (started fro rest), and so y = 1 t = y (3.00 ) = = 1.5 s t.00 s We can now substitute the acceleration back into our equation for M Proble 1 M R T M = ( ) = 4.0 kg 9.8 s 1.5 s 1.5 s = 46.5 kg

5 (cj18-037) The figure shows a box of ass M=50.0 kg hung fro a horizontal bea. The bea has ass =30.0 kg and length l=1.40. The bea is attached to the wall on the left by a hinge. One the right, it is suspended by a rope fro a second wall, at an angle θ=143 as shown. The rope can withstand a axiu tension of 490 N. Find the axiu distance x fro the left that the box can be hung without breaking the rope. Proble T θ l P x M Solution (1/3): The syste is in static equilibriu. We will choose the hinge at P to be the axis of rotation and we now enuerate the forces and the associated torques (assue CCW to be +): (1) Force by the hinge: F hx in the x direction, F hy in the y direction they exert no torque about P. () Weight of the bea: in the y direction only, and exerts torque at center of bea τ c = l sin 90 = l (the vector fro the hinge to the center-of-ass of bea points right, and the gravitational force points down, which is 90 in the CW (negative) direction (3) Weight of the box: M in the y direction only, and exerts torque at right end of bea, and like the weight of the bea, it is also directed at 90. It acts at a distance x fro the hinge. τ C = xm sin 90 = xm

6 (cj18-037) The figure shows a box of ass M=50.0 kg hung fro a horizontal bea. The bea has ass =30.0 kg and length l=1.40. The bea is attached to the wall on the left by a hinge. One the right, it is suspended by a rope fro a second wall, at an angle θ=143 as shown. The rope can withstand a axiu tension of 490 N. Find the axiu distance x fro the left that the box can be hung without breaking the rope. Proble T θ l P x M Solution (/3): (4) Tension force of the cable (whose agnitude we set to be the axiu value of 490 N). The force is directed at φ = 180 θ = = 37, to the right and above the horizontal: the coponents are T cos 37 in the x, and T sin 37 in the y direction. The direction fro the radial vector to the tension is also φ = +37 (at right end of bea) τ T = lt sin 37 Now we set the su of force coponents in x and in y directions to zero: F hx + T cos 37 = 0 1 F hx M + T sin 37 = 0 Setting su of the torques to zero: τ c + τ C + τ T = 0 l xm + lt sin 37 = 0 (3) Solving for x fro equation (3) we get: l xm + lt sin 37 = 0 xm = lt sin 37 l

7 (cj18-037) The figure shows a box of ass M=50 kg hung fro a horizontal bea. The bea has ass =30 kg and length l=1.4. The bea is attached to the wall on the left by a hinge. One the right, it is suspended by a rope fro a second wall, at an angle θ=143 as shown. The rope can withstand a axiu tension of 490 N. Find the axiu distance x fro the left that the box can be hung without breaking the rope. Solution (3/3): And so we have (everything is known on the right side: Proble T θ l P x M x = lt sin 37 l M = N kg 9.8 s 50.0 kg 9.8 s x = 0.4 We actually did not need the force equations for this proble.

8 (hr11-053) In the figure (overhead view), a unifor rod of length l=0.500 and ass 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 fro above, the bullet s path akes 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 iediately after the collision, what is the bullet s speed v 0 just before ipact? Proble 3 l θ Solution (1/3): Consider the bullet + rod to be a syste (this is actually necessary because the bullet becoes 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 syste, so linear oentu 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 oentu L is conserved. Before collision: the rod was at rest: no contribution to total v. The bullet has angular oentu given by L b = r p b. In this case r points along the rod fro the rotation axis to the end struck by the bullet, i.e. to the right. The oentu p b of the bullet points up and to the right, so rotating fro r to p b rotates by angle φ = +θ = +60. And so L b points out of the page, or we say L b = +rp b sin 60 is positive (i.e. CCW)

9 (hr11-053) In the figure (overhead view), a unifor rod of length l=0.500 and ass 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 fro above, the bullet s path akes 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 iediately after the collision, what is the bullet s speed v 0 just before ipact? Solution (/3): Proble 3 Alternate forula for angular oentu of bullet before collision: L b = +v. It is positive because the bullet isses the axis to the right (traveling in the CCW sense around the axis). Here v = v 0, and is the iniu distance of approach, which is shown in the diagra added on this page. By trigonoetry: = l sin θ. And so the total angular oentu before the collision is given by: l L i = L b = v 0 = v 0 sin 60 After the collision: The bullet and the rod becoe a syste with cobined oent-ofinertia given by I = I ror + r = 1 1 Ml + l = M l = 1 1 M + 3 l And as the syste rotates at ω f =10 rad/s after the collision, the final angular oentu is: L f = Iω f = 1 1 M + 3 l ω f l

10 (hr11-053) In the figure (overhead view), a unifor rod of length l=0.500 and ass 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 fro above, the bullet s path akes 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 iediately after the collision, what is the bullet s speed v 0 just before ipact? Solution (3/3): Angular Moentu is conserved, so we set L i = L f : Solving for v 0 then gives l 1 v 0 sin 60 = 1 M + 3 l ω f Proble 3 v 0 = l sin M + 3 l ω f = M + 3 lω f 6 sin 60 = 4.00 kg kg rad s kg 3 = s

11 (sj10-09) The figure shows a ca, which is a circular disk rotating on a shaft that does not pass through the center of the disk. For this ca, a unifor solid cylinder of radius R is first achined. Then an off-center hole of radius R/ is drilled, parallel to the axis of the cylinder, and centered at a point a distance R/ fro the center of the cylinder. The ca, of ass M, is then slipped onto the circular shaft and welded into place. Find the kinetic energy of the ca when it is rotating with angular speed ω about the axis of the shaft? Proble 4 P. 1/3 Consider a piece that is a full cylinder consisting of the ca plus a piece that exactly plus the hole. The area of the hole is ¼ that of the full disk, and that of the ca is ¾ that the full disk. So the ass of the holem hooo is related to the ass of the ca M by M hooo = M 3 And the ass of the full disk (FD) is therefore M FF = 4M 3. For this proble we will assue that the full disk is the su of the ca and the hole, which also works for calculating the oent of inertia about the center of the hole (shaft): I FF = I ccc + I hooo

12 (sj10-09) The figure shows a ca, which is a circular disk rotating on a shaft that does not pass through the center of the disk. For this ca, a unifor solid cylinder of radius R is first achined. Then an off-center hole of radius R/ is drilled, parallel to the axis of the cylinder, and centered at a point a distance R/ fro the center of the cylinder. The ca, of ass M, is then slipped onto the circular shaft and welded into place. Find the kinetic energy of the ca when it is rotating with angular speed ω about the axis of the shaft? Proble 4 P. /3 But since the oent-of-inertia of a cylinder (both full disk and hole) is straight-forward, we will then calculate I ccc by I ccc = I FF I hooo R hooo =R M hooo = M 3 We start with the hole: it is a unifor disk and rotates about its center: I hooo = 1 M hooor hooo = 1 M 3 R 1 = 4 MR

Proble 4 (sj10-09) The figure shows a ca, which is a circular disk rotating on a shaft that does not pass through the center of the disk. For this ca, a unifor solid cylinder of radius R is first P.

13 Proble 4 (sj10-09) The figure shows a ca, which is a circular disk rotating on a shaft that does not pass through the center of the disk. For this ca, a unifor solid cylinder of radius R is first P. 3/3 achined. Then an off-center hole of radius R/ is drilled, parallel to the axis of the cylinder, and centered at a point a distance R/ fro the center of the cylinder. The ca, of ass M, is then slipped onto the circular shaft and welded into place. Find the kinetic energy of the ca when it is rotating with angular speed ω about the axis of the shaft? I ccc = I FF I hooo R FF = R D M FF = 4M 3 Full Disk: by Parallel Axis Theore: noting the actual axis of rotation is a distance D = R/ fro the center of the full disk. I FF = I FF(CC) + M FF D = 1 M FFR R + M FF = 1 4M 3 R + 4M 3 R 4 = MR = MR I ccc = I FF I hooo I ccc = MR 1 4 MR = Rotating at angular velocity ω K = 1 I cccω = MR ω = 3 48 MR ω 3 4 MR

NAME NUMBER SEC. PHYCS 101 SUMMER 2001/2002 FINAL EXAME:24/8/2002. PART(I) 25% PART(II) 15% PART(III)/Lab 8% ( ) 2 Q2 Q3 Total 40%

NAME NUMBER SEC. PHYCS 101 SUMMER 2001/2002 FINAL EXAME:24/8/2002. PART(I) 25% PART(II) 15% PART(III)/Lab 8% ( ) 2 Q2 Q3 Total 40% NAME NUMER SEC. PHYCS 101 SUMMER 2001/2002 FINAL EXAME:24/8/2002 PART(I) 25% PART(II) 15% PART(III)/Lab 8% ( ) 2.5 Q1 ( ) 2 Q2 Q3 Total 40% Use the followings: Magnitude of acceleration due to gravity