Pendulum. A small rock with mass 0.30 kg is fastened to a massless string withle

Question

Pendulum. A small rock with mass 0.30 kg is fastened to a massless string withlength 0.86 m to form a pendulum. Thependulum is swinging so as to make a maximum angle of 45° withthe vertical. Air resistance is negligible.

(a) What is the speed of the rock when the string passes throughthe vertical position?


(b)What is the tension in the string as it passes through thevertical?

Explanation / Answer

mgcos45= .3(9.8)(cos45)= 1.909 N As the sum of all forces equals the mass times acceleration f=ma Conservation of energy: 1/2 m v^2 = mgl (1-cos 45) ==> v =(2gl(1-cos 45)) = 2.22 m/s in this case you have a radial acceleration so you need to solvefor it using v^2/r where v is velocity and r is the radius so you have 2.22^2.22/.86=5.74 know that F(tension)-F(gravity)=5.74*m F(tension)=1.722+(.3)(9.8)=4.662 N

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