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A ball of mass 0.300 kg is whirled in a horizontal circle at the end of a light string 75.0 cm long at a speed of 2.00 m/s. If the string is lengthened to 2.00 m while the ball is being twirled, then the speed of the ball will now be?

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Final answer:

To find the speed of the ball after the string is lengthened, we can use the concept of conservation of angular momentum. By setting the initial and final angular momentum equal and solving for the final angular velocity, we fin that the speed of the ball will be approximately 0.5625 m/s.

Step-by-step explanation:

To find the speed of the ball after the string is lengthened, we can use the concept of conservation of angular momentum. Angular momentum is given by the equation L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity. Since the string is lengthened, the moment of inertia will increase, but the angular momentum must remain constant. Therefore, we can set the initial angular momentum equal to the final angular momentum and solve for the final angular velocity.

Initial angular momentum: Li = Iiωi

Final angular momentum: Lf = Ifωf

Since the ball is whirled in a horizontal circle, we can assume the moment of inertia is proportional to the mass of the ball and the square of its distance from the axis of rotation. So, Ii = miri2 and If = mfrf2

Substituting these values into the equation for angular momentum and solving for ωf, we get:

ωf = ωi * (ri/rf)2

Plugging in the given values, we have ωi = 2.00 m/s and ri = 75.0 cm = 0.75 m. When the string is lengthened to 2.00 m, rf = 2.00 m. Substituting these values, we can calculate the final angular velocity:

ωf = 2.00 m/s * (0.75 m/2.00 m)2 = 0.5625 m/s

Therefore, the speed of the ball after the string is lengthened will be approximately 0.5625 m/s.

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