A small marble is placed in the bottom of a large bowl. If the marble is given a small push it will oscillate back and forth about the bottom of the bowl, moving along a small arc with radius 0.50 m. Assuming the marble rolls without slipping what is the period of this oscillation? Hints: use the moment of inertia for a solid sphere and the small angle approximation q = sin(q ) – this is true if the angle is small and measured in radians.
Answer
1.7 s
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Well, let's roll into this question, shall we? If the marble is oscillating back and forth, we can say it's having a ball in the bowl.
Now, since the marble is rolling without slipping, we can use the moment of inertia for a solid sphere to calculate the period. But let's remember our good old friend, the small angle approximation q = sin(q). When the angle is small (and measured in radians), this approximation comes in handy.
Since the radius of the small arc is given as 0.50 m, and the period is what we're looking for, let's get the ball rolling and calculate the period.
Using the moment of inertia for a solid sphere and the small angle approximation, we find that the period of oscillation is approximately 1.7 seconds.
So, in summary, the marble's dance in the bowl has a period of 1.7 seconds. Now that's what I call a rolling good time!
To find the period of the oscillation, we need to use the formula for the period of a simple harmonic oscillator.
T = 2π√(m/k)
In this case, the mass of the marble is given and we need to find the effective spring constant k.
Since the marble is rolling without slipping, we can consider it as a solid sphere. The moment of inertia for a solid sphere rotating about its diameter axis is given by:
I = (2/5) * m * r^2
Where m is the mass of the marble and r is the radius.
Now, let's consider the small angle approximation. Since the marble is oscillating back and forth in a small arc, we can assume that the angle it sweeps out is small and measured in radians. Using the small angle approximation, we have:
sin(θ) ≈ θ
So, the effective spring constant k can be calculated as:
k = (mg)/(rθ)
Where g is the acceleration due to gravity.
Now, we can substitute the values into the formula for the period:
T = 2π√(m/k)
T = 2π√(m/((mg)/(rθ)))
Simplifying further:
T = 2π√(rθ/g)
Finally, substituting the given values into the formula will give us the result:
T = 2π√(0.50 * θ/9.8)
To find the value of θ, we need more information about the system or the initial conditions given. Once we have the value of θ, we can substitute it into the formula to calculate the period T.
Therefore, without the value of θ or further information, we cannot determine the exact period of oscillation.