Two 30 kg twins climb onto a lightweight seesaw: Sal sits 1.5 m from the pivot,

Question

Two 30 kg twins climb onto a lightweight seesaw: Sal sits 1.5 m from the pivot, and Pat sits 1.0 m from the pivot on the same side as Sal.

(A) How much torque do the two kids exert about the pivot, assuming the board is horizontal?
(B) How large is the moment of inertia of the kids about the pivot?  
(C) Their 80 kg dad climbs on the seesaw on the other side.   Where should he be to balance them?
(D) If dad then abruptly slides off the seesaw, what is its initial angular acceleration?  
(E) What is the downward linear acceleration of the board under Sal?

how do you get angular acceleration without a radius and how do you find moment of inertia?? and how do you get acceleration without time or velocity?

Explanation / Answer

A) net torque = 30*9.81 (1.5-1) Nm =147.15 Nm

B) 30* 1.5*1.5+30*1*1 = 97.5 Kgm2

C)30*1.5=30*1+80*x

x = 0.1875 meter

D) torque = I * alpha

alpha = 147.15/97.5 =1.50923076923 rad/s2

E) linear acc = 1.5 * alpha = 2.26384615384

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