Question A pendulum in (Figure 1) has a swinging motion such that Part A \( \theta=[-0.101 \sin (4.15 t)+0.3 \cos (4.15 t)] \) rad, where \( t \) is in seconds, and the arguments for the sine and cosine are in radians. Determine the magnitude of the velocity of the bob when \( \theta=0^{\circ} \). Assume \( L=0.2 \mathrm{~m} \). Express your answer to three A pendulum in (Figure 1) has a swinging motion such that Part A θ=[−0.101sin(4.15t)+0.3cos(4.15t)] rad, where t is in seconds, and the arguments for the sine and cosine are in radians. Determine the magnitude of the velocity of the bob when θ=0∘. Assume L=0.2 m. Express your answer to three significant figures and include the appropriate units. Part B Determine the magnitude of the acceleration of the bob when θ=0∘. Express your answer to three significant figures and include the appropriate units.

EAU0RT The Asker · Mechanical Engineering

Transcribed Image Text: A pendulum in (Figure 1) has a swinging motion such that Part A θ=[−0.101sin(4.15t)+0.3cos(4.15t)] rad, where t is in seconds, and the arguments for the sine and cosine are in radians. Determine the magnitude of the velocity of the bob when θ=0∘. Assume L=0.2 m. Express your answer to three significant figures and include the appropriate units. Part B Determine the magnitude of the acceleration of the bob when θ=0∘. Express your answer to three significant figures and include the appropriate units.

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Here I solved the above problem with detailed solution as below. I upload the images of above mentioned problem with step by step solution. This is the problem of Engineering Mechanics. => Solution:rarr Given dota:theta=[-0.101 sin(4.15 t)+0.3 cos(4.25 t)] rod whats, t is in second.L=0.2mrarr To find:Part A) Mognitude of velocitf of the bob when theta=0^(@)Port B) Mognitide of acceleration of bob when E=0^(@)rarr calculation:Finding the time of theta=0^(@)=0rad.{:[:.0=-0.101 sin(4.15 t)+0.3 cos(4.15 t)],[0.101 sin(4.15 t)=0.3 cos(4.15 t)],[:.tan(4.15 t)=(0.3)/(0.101)=2.97],[:.4.15 t=tan^(-1)(2.97)=71.39],[:.t=(71.39)/(4.15)=17.2],[:.t=17.2sec" ot "theta=0^(@)" radian "]:}{:[" Nots, "theta=0^(@)" (Aegree) "],[t=(17.2 xx pi)/(180)=0.3sec.],[:.t=0.3sec" at "theta=0^(@)" (Agree) "]:}Dest A) Angulor velo atf, omega=(d theta)/(dt){:[:.omega=(d theta)/(d theta)=(d)/(dt)[-0.101 sin(4.15 t)+0.3 cos(4.15 t)]],[:.(d theta)/(dt)=-0.101 cos(4.25 t)xx4.15+0.3[-sin(4.15 t)xx4.25]],[:.(d theta)/(dt)=4.15[-0.101 cos(4.25 t)-0.3 sin(4.15 t)]],[" Noke, "At","t=0.3sec". "],[(d theta)/(dt)]_(t=0.3)=4.15[-0.101 cos(4.25 xx0.3)-0.3 sin(4.25 xx0.3)]],[=4.15[-0.101 cos(1.245)-0.3 sin(1.245)]],[=4.15[-0.1009-0.006]],[=4.15(-0.1069)], ... See the full answer