Problem Set 11

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SUNY Empire State College *

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1200

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Mechanical Engineering

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Feb 20, 2024

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docx

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Uploaded by DeaconMosquitoMaster1223

Problem Set # 11 – Problems 2, 12, 24, 40, 42, 64 2. The gardening tool is used to pull weeds. If a 1.23-N ̇ m torque is required to pull a given weed, what force did the weed exert on the tool? For this, force needs to be solved. Force = torque/r Torque = 1.23 -N *m r= 0.40m (1.23N*m)/.040m = 31N 12. When the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 3.0 revolutions. If the CD has a radius of 6.0 cm and a mass of 17 g, what is the torque exerted on it? Given: Radius = 0.060 m Acceleration of CD = 450 rev/min in 3.0 revolutions M = 0.017g First part to solve would set up as: (450rev/min x 2pi rad/rev x 1min/60sec) ^2 - 0^2 / 2 x 3rev x 2pi rad/rev = 59rad/sec^2 Now we solve for the second part which is Inertia = 0.5 MR^2 0.5 (0.017kg) (0.060m) ^2 = 3.1×10^−5 kg * m^2 Now to solve for Torque it would follow as: (3.1×10^−5 kg * m^2) x (59rad/sec^2) = 0.0018 N*m is the torque exerted on the CD.

24. To determine the location of her center of mass, a physics student lies on a lightweight plank supported by two scales 2.50 m apart. If the left scale reads 435 N, and the right scale reads 183 N, find (a) the student’s mass and (b) the distance from the student’s head to her center of mass. Given: F1 = 435 N F2 = 183 N L = 2.5 m To solve mass, it would follow as: m x g = F1 + F2  m = F1 + F2 / g m = 435 N + 183 N / 9.81m/s^2 m = 62.99  63 kg is the student’s mass. b) to find the center of mass, the principles of torque would be used as: T = L x F2 – x x W = 0  x = L x F2 / m x g x = 2.5m x 183 N / 63kg x 9.81m/s^2 x = 0.74m is the distance from student’s head to her center of mass

40. A 2.85-kg bucket is attached to a rope wrapped around a disk-shaped pulley of radius 0.121 m and mass 0.742 kg. If the bucket is allowed to fall, (a) what is its linear acceleration? (b) What is the angular acceleration of the pulley? (c) How far does the bucket drop in 1.50 s? Given: Mass (bucket): 2.85 kg Radius (pulley): 0.121 m Mass of pulley: 0.742 kg gravity = 9.81 m/s^2 Inertia = 0.5 mr^2 a) Solving linear acceleration would follow as: a = - 9.81 / 1.130175 = 8.68 m/s^2 linear acceleration is 8.68 m/s^2 b) Solving Angular acceleration would follow as: a / r = 8.68 / 0.121 = 71.74 rad/s^2 angular acceleration of the pulley is 71.74 rad/s^2 c) To solve Acceleration equation of motion, it would follow as: s = 0 + 0.5 x 8.68 x 1.5^2 = 9.765 m In 1.50 s = bucket drops at 9.765 m 42. You pull downward with a force of 28 N on a rope that passes over a disk-shaped pulley of mass 1.2 kg and radius 0.075 m. The other end of the rope is attached to a 0.67-kg mass. (a) Is the tension in the rope the same on both sides of the pulley? If not, which side has the greater tension? (b) Find the tension in the rope on both sides of the pulley. List of Given Variables: 1. Force applied (downward) = 28 N which is T1 2. Mass of the pulley (m ) = 1.2 kg ₁ 3. Radius of the pulley (r) = 0.075 m 4. Mass attached to the other end of the rope (m2) = 0.67 kg (a)No, the tension in the rope on the other end of the rope accelerates the hanging mass, but the tension on the opposite side both imparts angular acceleration to the pulley and

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accelerates the hanging mass. Therefore, the rope on the opposite side (my side) where it is pulled and does not have the hanging mass has the greater tension. (b) To solve for T2 we would follow as: T2 = (0.67kg) [(2(28N) + (1.2kg) (9.81 m/s^2)] / 2 (0.67kg) + 1.2kg which would = 18N 64. A popular make of dental drill can operate at a speed of 42,500 rpm while producing a torque of 3.68 oz ·in. What is the power output of this drill? Give your answer in watts. Given:  Angular speed = 42,500 rpm  Torque = 3.68 oz ·in Torque= 3.68 oz ·in x(1lb/16oz) x (4.45N/1lb) x (1m/39.4in) = .0260N*m Angular speed = 42500rev/min x (2pirad/rev) x (1min/60sec) = 4450rad/s Now we solve for P which is A x T: (0.0260N*m) (4450 rad/s) = 116 W is the power output for this drill.