The diagram shows a thin rod of uniform mass distribution pivoted about one end by a pin passing through that point. The mass of the rod is 0.600 kg and its length is 2.00 m. When the rod is released from its horizontal position, it swings down to the vertical position as shown.(a) Determine the speed of its center of gravity at its lowest position.
Question
The diagram shows a thin rod of uniform mass distribution pivoted about one end by a pin passing through that point. The mass of the rod is 0.600 kg and its length is 2.00 m. When the rod is released from its horizontal position, it swings down to the vertical position as shown.(a) Determine the speed of its center of gravity at its lowest position.
Solution
To solve this problem, we can use the principle of conservation of energy. This principle states that the total energy in a closed system remains constant. In this case, the system is the swinging rod.
Step 1: Identify the initial and final states of the system. The initial state is when the rod is horizontal and at rest. The final state is when the rod is vertical and moving.
Step 2: Calculate the initial potential energy (PEi). The potential energy is given by the formula PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. In the initial state, the center of gravity of the rod is at a height of 1.00 m (half the length of the rod) from the pivot point. So, PEi = (0.600 kg)(9.81 m/s^2)(1.00 m) = 5.886 J.
Step 3: Calculate the final potential energy (PEf). In the final state, the center of gravity of the rod is at a height of 0 m from the pivot point. So, PEf = 0 J.
Step 4: Apply the conservation of energy. The initial total energy is equal to the final total energy. So, PEi + KEi = PEf + KEf, where KE is the kinetic energy. Since the rod starts from rest, KEi = 0. So, 5.886 J = 0 J + KEf. Therefore, KEf = 5.886 J.
Step 5: Calculate the final speed. The kinetic energy is given by the formula KE = 0.5mv^2, where m is the mass and v is the speed. So, v = sqrt((2KEf)/m) = sqrt((25.886 J)/(0.600 kg)) = 4.90 m/s.
So, the speed of the center of gravity of the rod at its lowest position is 4.90 m/s.
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