the velocity function v(t) (5 points) the position function x(t) (5 points) a. b
ID: 1782757 • Letter: T
Question
the velocity function v(t) (5 points) the position function x(t) (5 points) a. b. Rotation and Torque (20 points) An 81.2 cm rod of uniform density is placed on a fulcrum. 1. If a mass of 4 kg is placed 20 cm away from the fulcrum, which itself is 20.6 cm away from the 2. Using the mass of the meter stick (stia) found in question 1, determine the unknown mass (m) 3. Why was the supporting force exerted on the meter stick by the sharp edge (the one placed in COG, what will be the mass of the meter stick? (8 points) if there is a 4 kg mass (M) placed 22 cm from the fulcrum, the fulcrum is 18.6 cm away from the COG and the position of the unknown mass is at the 81 cm mark. (8 points) the fulcrum) not considered in your calculations? (4 points) Bonus: A meter stick is pivoted at its 50 cm mark but does not balance because of non-uniform ties in its material that cause its center of gravity to be displaced from its geometrical center. However when weights of 150 g and 300 g are placed at the 10 cm and 75 cm marks, respectively, balance obtained. The weights are then interchanged and 'balance is again obtained by shifting the pivo point to the 43 cm mark. Find the mass of the meter stick and the location of its center of gravit (10 points)Explanation / Answer
net torque about the fulcrum = 0
mg*l1 = Mg*l2
m*l1 = M*l2
4*20 = M*20.6
Mass of rod = 3.88 kg
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2)
position of fulcrum = 40.6 - 18.6 = 22 cm
mass m1 = 4 kg is at 0 cm mark
m1*l1 = M*l + m2*l2
distance of m1 from fulcrum l1 = 22 cm
distance of COG from fulcrum l = 18.6 cm
distance of unknown mass m2 from fulcrum l2 = 81 - 22 = 59 cm
4*22 = 3.88*18.6 + (m2*59)
m2 = 0.27 kg
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3)
the torque exerted by supporting force on fulcrum = 0
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