Question

Fino THE Max HUM TARGETAL AROA Nora ACCELERATIONS F A SinPLE PEDULUR. - 30 CLEGHAT & zin, n = 10 kg e

Answer 1

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solution: given data Omax = 30 l o L = 2m m. lokg find (Pendulum is maximum tangantial accetration Normal acceleration, 23 by the free free body diagram of the Pendulum LU String mg here, there are two Component of the say that na Weight We can (mg) or of the Pendulum, 1) Perpendicular to the string along the string 2)

Newtonk 2nd law mg mg cose along the string is named as Normal comporrent; normal acceleration is an ] mgcoso is equate to mgsino ong cosco is equivalent to man Where an = 9 m/s2 (1 D perpendicular to the string is named Tangential components Tengential acceleration is cat) mg sina is equivalent to mat Where 2. cu = 9 sind m/s2

g Cosomax so ; 1) Maximum Normal acceleration (an) max 19 9.88 Cos 30 (an man 13 8.4870 m/s2 maximum tangential acceleration (at) max s sin @ max gos 8x Sin 30 at max =4.9 mis?

→ for maximum normal acceleration lcoso L Ice- deose) A from B to A there will be increase in Speed and mean position A Speed normal is maximum acceleration is maximum. so that conservation between so; by Energy A and B decrease in potential Energy increase in kinetic Energy to mgl- mgl caso = Į mv2 1 ge (1-cosol) lu2 2 2 V - 2 x 9.8 x 2cl-COS 30) 5.2518

normal cecceleration is maximum ( Qmax : V 2 d 8.2518 2. 625 m/seel 2 CAN max = 2.625 m/s²