Question

1. A pendulum consists of a 25-kg bob and a massless rod 2.8 meters long. The bob is raised to the horizontal position and released from rest. Find the tension in the rod when the rod is vertical. | 2.5m o* K +U = k' + mgh = / 2 mv² +0 Convention Ta gh = 1 / 2 = √29h v 2(.8) (2.8) :: v: 7.4 m/s Fnet a ma T-mg = m.it T - (25)(9.8) = 25.74 :. T = 733.93 N 1

PLEASE SOLVE TO GET THE VELOCITY OF THE ***BOB*** (not the rod) when the rod is vertical
thank you

Answer 1

th=2. 8m. m = mass of bob = 25 kg Energy of Kit 0 Rob at position I a Energy of Bob at position 2 K2_t Uz K & U 8 K U2 = Kinetic_Energy Of Bob at position 2 = Potential Energy of Bob at position 1&2. of Bob = omis] ki = o I [.: velocity U = mgh - = 25 x 9.8 x 2.8 J [v = velocity of Bob at 2] Ky U₂ = I mu? = o J 0 + (25* 9.8 x 2.8) = 64 x 25% v2) to = > - V2 v 2x 2.8 *9.8 7.41 mis.] =

Now when bob a is at position 2 (ood is _ vertical). Two Forces are acting on it - One Force is downwards due to its mass - 2nc Farce is also down coards due to its velocity! in arcular motion... Tension, I = mg + mv ² = m(8 + x) = 25 ( 9.8 + 54.88 - [ 735 N] 2.8
The solution is same as the one you provided with the question. We just need to conserve energy of Bob on 2 positions here. I tried to explain it but still if you are not able to understand or have any kind of doubt in the solution, feel free to ask.