Question

A simple pendulum has a mass of 0.650 kg and a length of 7.00 m. It is displaced through an angle of 8.0° and then released. Using the analysis model of a particle in simple harmonic motion, calculate the following. (Give your answer to the thousandths place.) (a) What is the maximum speed of the bob? m/s 4.0 (b) What is the maximum angular acceleration of the bob? | rad/s² 4.0 (c) What is the maximum restoring force of the bob? 4.0 N 4.0 (d) Solve parts (a)through (c) by using other analysis models. (Hint: you may need to use separate analysis models for each part.) maximum speed m/s maximum angular acceleration 49) rad/s2 maximum restoring force 4.0 N (e) Compare the answers.

Please answer with the correct amount of significant figures. Thank you!!

Answer 1

given Mass of pendulum m= 0.650 kg length of pendulum L 7m Maximum angle 0 = 8° .T L Cose mg sino Yong Cose mg 0 IS The formula to calculate amplitude A = LO where L is the length of pendulum &o ů the displaced angle Substitute I go A = X 8XT 180 A = 0.977 0.98m Angular frequency w is geven by w Ja/ where is the accůleration due to gravity substituting for g&L, we get

o = 9.8 =1.4 Fm w 1.18 vad /s a) Maximum Speed max Aw Vmase = (0.98)x (1.18) Vmax = 1.16 m/s Maximum angular acculeration a amax = Aue 2 20 amax (0.98) (1.18) (1.18)2 = 1.36m/52 Lmax amao L 0.19 vad/s2 7 c) Maximum restoring force ma F = 6.65kg) (136) E 0.88 N

kinetic energy Maximum speed mgh - ₂ mva ié potential energy Vmax 294 (1-cose) since b= l (1-Cose), From friguve Vmax of axa 2x9.8)* 7m* (1– Cos 8) re Vmax 1.34 Vman 1.16 W/ Maximum angular accileration Τα mgLsino I is the moment of cinertia. I = mL? mgh c MLL sino a is the angular accilevation 9 은 Sino lmane mgL sine ML

Iman (918) + sin (8) 7 Sman 0.19 rad/s2 Fmax mg sina Fmax (0.65) x (9.8) * sin() Fmax 0.88N