A uniform rod of mass 2 kg and length 1.41 m is hinged at its top end and allowed to swing in the vertical plane of the paper. It is initially stationary in a vertical position. A clay ball of mass 0.25 kg is propelled towards its lower end at a speed v. The ball sticks to the end of the rod on impact. What ball speed v is needed for the rod and the clay ball to rise to an angle of 90 degrees before stopping and falling back?
I first used the conservation of angular momentum (mvr = Iw) to get that v = 5.17w.
Then I used the conservation of energgy theorem (0.5Iw^2(initial) + mgh(initial) = 0.5Iw^2(final) + mgh(final)). However I get my answer wrong. I cannot understand why there is gravitational potential energy at the initial position because the center of mass has not moved yet. And does the center of mass not change when the clay ball sticks to the rod.
Please help?
Thank you.
To solve this problem, let's break it down step by step:
Step 1: Conservation of Angular Momentum
Using the conservation of angular momentum, we can write the equation mvr = Iw. Here, m is the mass of the clay ball (0.25 kg), v is the velocity at which the ball is launched, r is the length of the rod (1.41 m), I is the moment of inertia of the rod, and w is the angular velocity of the system.
Step 2: Moment of Inertia
The moment of inertia, I, for a uniform rod rotating about its end is given by I = (1/3) * m * L^2, where m is the mass of the rod (2 kg) and L is the length of the rod (1.41 m). Substituting these values, we get I = (1/3) * 2 kg * (1.41 m)^2.
Step 3: Solving for Angular Velocity
Now that we have the expression for I, we can substitute it into the angular momentum equation to solve for w. So, mvr = Iw becomes 0.25 kg * v * 1.41 m = (1/3) * 2 kg * (1.41 m)^2 * w.
Simplifying the equation, we get: v = (1/3) * 2 kg * (1.41 m) * w / 0.25 kg.
Step 4: Conservation of Energy
Next, we can use the conservation of energy theorem to find the value of w. The equation is given by: 0.5Iw^2(initial) + mgh(initial) = 0.5Iw^2(final) + mgh(final).
At the initial vertical position, there is no gravitational potential energy (hence mgh(initial) = 0), but there is initial kinetic energy (0.5Iw^2(initial)). So, the equation simplifies to: 0.5Iw^2(initial) = 0.5Iw^2(final) + mgh(final).
Step 5: Finding the Final Angle
Since the final angle is 90 degrees, we know that h(final) = L/2, where L is the length of the rod. Substituting these values, we get 0.5Iw^2(initial) = 0.5Iw^2(final) + mg(L/2).
Step 6: Solving for Angular Velocity Again
Now, let's substitute the expression for I and solve for w. After simplification, the equation becomes: 0.5 * (1/3) * 2 kg * (1.41 m)^2 * w^2 = 0.5 * (1/3) * 2 kg * (1.41 m)^2 * w^2 + (0.25 kg * 9.8 m/s^2 * (1.41 m)/2).
Step 7: Simplification
We notice that the term (1/3) * 2 kg * (1.41 m)^2 * w^2 cancels out, leaving us with 0 = (0.25 kg * 9.8 m/s^2 * (1.41 m)/2).
Step 8: Solving for the Final Answer
Finally, we can solve the equation to find the value of w, which represents the angular velocity. By simplifying the equation, we find that 0 = 0.25 * 9.8 * (1.41/2).
Therefore, w = 0 rad/s.
Step 9: Finding Ball Speed v
Now that we know w = 0 rad/s, we can substitute this value back into the equation v = (1/3) * 2 kg * (1.41 m) * w / 0.25 kg.
v = (1/3) * 2 kg * (1.41 m) * 0 rad/s / 0.25 kg.
v = 0 m/s.
So, the ball speed needed for the rod and clay ball to rise to an angle of 90 degrees before stopping and falling back is 0 m/s.