A ruler is resting on the edge of a ledge, with 2/9 of its length hanging over t

Question

A ruler is resting on the edge of a ledge, with 2/9 of its length hanging over the edge. A single beetle lands on the end of the ruler hanging over the edge, and the ruler begins to tip. A moment later, a second, identical beetle lands on the other end of the ruler, which results in the ruler coming momentarily to rest at 62.1deg from the horizontal. If the mass of each bug is 3.26 times the mass of the ruler and the ruler is 18.7 cm long, what is the magnitude of the angular acceleration of the ruler at the instant described?

Explanation / Answer

on hanging edge

m1 = 3.26 M

x1 = 2L/9

opposite edge

m2 = 3.26 M

x2 = 7L/9

net torque = M*g*L/2*cos62.1 + m2*g*x2*cos62.1 - m1*x1*cos62.1

net torque = M*g*L/2*cos62.1 + m2*g*7L/9*cos62.1 - m1*2L/9*cos62.1

net torque = [ (0.5) + (3.26*(7/9))   - (3.26*(2/9)) ]* M*g*L

net torque = 2.311*M*L*g

torque = I*alfa

I = (1/12)M*L^2 + M*(L/2-2L/9)^2 + m1*x1^2+m2*x2^2

I = ( (1/12) + 0.077 + (2/9)^2 + (7/9)^2 )M*L^2

I = 0.815*M*L^2

2.311*M*L*g = 0.815*M*L^2*alfa

2.311*9.8 = 0.815*0.187*alpha

alpha = 148.6 rad/s^2 <<-----answer

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