Physics 8 Monday, October 28, 2013

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1 Physics 8 Monday, October 28, 2013 Turn in HW8 today. I ll make them less difficult in the future! Rotation is a hard topic. And these were hard problems. HW9 (due Friday) is 7 conceptual + 8 calculation problems. Of the 8 calculation problems, 4 or 5 are from Chapter 11, and 3 or 4 are from Chapter 12. Reload the online-response page in your web browser. Read Chapter G9 (Giancoli s chapter 9) for Wednesday: Static equilibrium, elasticity, fracture. It relates closely to the topic of torque. We have the next 2 3 weeks to understand & use torque. Quiz Weds: one easy problem from HW7. Look over your graded HW7, which was handed back last Friday.

2 reminder: correspondence between linear and rotational motion position rotational coordinate velocity acceleration r = (x, y) v = (v x, v y ) = d r dt a = (a x, a y ) = d v dt if a x is constant then: v x,f = v x,i + a x t x f = x i + v x,i t a xt 2 v 2 x,f = v 2 x,i + 2a x x ϑ = s/r rotational velocity ω = dϑ dt rotational acceleration α = dω dt if α is constant then: ω f = ω i + αt ϑ f = ϑ i + ω i t αt2 ω 2 f = ω 2 i + 2α ϑ

3 Torque: the rotational analogue of force Just as an unbalanced force causes linear acceleration F = m a an unbalanced torque causes rotational acceleration τ = I α Torque is (lever arm) (force) τ = r F where r is the perpendicular distance from the rotation axis to the line-of-action of the force.

4 position rotational coordinate r = (x, y) ϑ = s/r velocity acceleration force v = (v x, v y ) = d r dt a = (a x, a y ) = d v dt F = m a rotational velocity ω = dϑ dt rotational acceleration torque α = dω dt τ = I α τ = r F

5 I wind a string around a coffee can of radius R = 0.05 m. (That s 5 cm.) Friction prevents the string from slipping. I apply a tension T = 20 N to the free end of the string. The free end of the string is tangent to the coffee can, so that the radial direction is perpendicular to the force direction. What is the magnitude of the torque exerted by the string on the coffee can? (A) 1 N m (B) 2 N m (C) 5 N m (D) 10 N m (E) 20 N m

6 Suppose that the angular acceleration of the can is α = 2 s 2 when the string exerts a torque of 1 N m on the can. What would the angular acceleration of the can be if the string exerted a torque of 2 N m instead? (A) α = 0.5 s 2 (B) α = 1 s 2 (C) α = 2 s 2 (D) α = 4 s 2 (E) α = 5 s 2 (F) α = 10 s 2

7 I apply a force of 5.0 N at a perpendicular distance of 5 cm (r = 0.05 m) from this rotating wheel, and I observe some angular acceleration α. What force would I need to apply to this same wheel at r = 0.10 m (that s 10 cm) to get the same angular acceleration α? (A) F = 1.0 N (B) F = 2.5 N (C) F = 5.0 N (D) F = 10 N (E) F = 20 N

8 Suppose that I use the tension T in the string to apply a given torque τ = r T to this wheel, and it experiences a given angular acceleration α. Now I increase the rotational inertia I of the wheel and then apply the same torque. The new angular acceleration α new will be (A) larger: α new > α (B) the same: α new = α (C) smaller: α new < α

9 I want to apply to this meter stick two torques of the same magnitude and opposite sense, so that the stick has zero rotational acceleration. I apply one force of 5 N at a lever arm of 0.5 m. I want to apply an opposing force at a lever arm of 0.2 m, so that the second torque balances the first torque. How large must this second force be? (A) 1.0 N (B) 2.0 N (C) 12.5 N (D) 25 N

10 I want to apply to this meter stick two torques of the same magnitude and opposite sense, so that the stick has zero rotational acceleration. I apply one force of 10 N at a lever arm of 0.5 m. I tie a second string on the opposite end, 0.5 m from the pivot point. The second force is applied at a 45 angle w.r.t. the vertical. How large must this second force be? (A) 5 N (B) 7 N (C) 10 N (D) 14 N (E) 20 N

11 τ = r F = rf = rf sin θ rf = r F Four ways to get the magnitude of the torque (perpendicular component of distance) (force) (distance) (perpendicular component of force) (distance) (force) (sin θ between r and F ) use magnitude of vector product r F (a.k.a. cross product )

12 If the rod doesn t accelerate, what force does the scale read? (A) 1.0 N (B) 5.0 N (C) 7.1 N (D) 10 N (E) 14 N (F) 20 N

13 If the rod doesn t accelerate, what force does the scale read? (A) 1.0 N (B) 5.0 N (C) 7.1 N (D) 10 N (E) 14 N (F) 20 N

14 If the rod doesn t accelerate, what force does the scale read? (A) 1.0 N (B) 5.0 N (C) 7.1 N (D) 10 N (E) 14 N (F) 20 N

15 If the rod doesn t accelerate, what force does the scale read? (A) 1.0 N (B) 5.0 N (C) 7.1 N (D) 10 N (E) 14 N (F) 20 N

16 If the rod doesn t accelerate, what force does the scale read? (A) 1.0 N (D) 10 N (B) 5.0 N (E) 14 N (C) 7.1 N (F) 20 N

17 To tighten a bolt, I apply a force of magnitude F at different positions and angles. Which torque is largest?

18 To tighten a bolt, I apply a force of magnitude F at different positions and angles. Which torque is smallest?

19 I want to tighten a bolt to a torque of 1.0 newton-meter, but I don t have a torque wrench. I do have an ordinary wrench, a ruler, and a 1.0 kg mass tied to a string. How can I apply the correct torque to the bolt? (A) Orient the wrench horizontally and hang the mass at a distance 0.1 m from the axis of the bolt (B) Orient the wrench horizontally and hang the mass at a distance 1.0 m from the axis of the bolt

20 If the wrench is at 45 w.r.t. horizontal, will the 1.0 kg mass suspended at a distance 0.1 m along the wrench still exert a torque of 1.0 newton-meter on the bolt? (A) Yes. The force of gravity has not changed, and the distance has not changed. (B) No. The torque is now smaller about 0.71 newton-meter. (C) No. The torque is now larger about 1.4 newton-meter.

21 Physics 8 Monday, October 28, 2013 Turn in HW8 today. I ll make them less difficult in the future! Rotation is a hard topic. And these were hard problems. HW9 (due Friday) is 7 conceptual + 8 calculation problems. Of the 8 calculation problems, 4 or 5 are from Chapter 11, and 3 or 4 are from Chapter 12. Reload the online-response page in your web browser. Read Chapter G9 (Giancoli s chapter 9) for Wednesday: Static equilibrium, elasticity, fracture. It relates closely to the topic of torque. We have the next 2 3 weeks to understand & use torque. Quiz Weds: one easy problem from HW7. Look over your graded HW7, which was handed back last Friday.

Physics 8 Friday, October 20, 2017

Physics 8 Friday, October 20, 2017 HW06 is due Monday (instead of today), since we still have some rotation ideas to cover in class. Pick up the HW07 handout (due next Friday). It is mainly rotation, plus