Suppose you swing a ball of mass m in a vertical circle on a string of length L. As you probably know from experience, there is a minimum angular velocity ωmin you must maintain if you want the ball to complete the full circle without the string going slack at the top.(1) find an expression for the minimum angular velocity using only the terms (g) and (L).(2) evaluate the angular velocity in rpm for a 65 g ball tied to a 2.0-m-long string.

1) Let's use Newton's second law n for the top of the circle. In that case, when the speed is the same, the only force that is current is the body mass.

W = m a

Where the acceleration is centripetal

a = w² r

mg = m w2 r

w² = g / r

w = √ (g / r)

r = L

w = √ (g / L)

2) The angular velocity for 2.0 m rope

w = RA (9.8 / 2.0)

w = 2.21 rad / s

We reduce rad / s to rpm

w = 2.21 rad / s (1 rev / 2pi rad) (60s / 1 min)

w = 21.1 rpm

onyebuchinnaji onyebuchinnaji · Answer 2 · 2021-10-30T17:38:49+02:00

The minimum angular velocity of the object is [tex]\omega = \sqrt{\frac{g}{L} }[/tex].

The angular velocity of the ball at the given mass and length of the string is 21.14 rpm.

The given parameters;

Mass of the ball, = m
Length of the string, = L

The minimum angular velocity of the object is calculated as follows;

[tex]F= mg = m\omega^2 r\\\\g = \omega ^2 r\\\\\omega^2 = \frac{g}{r} \\\\\omega =\sqrt{\frac{g}{r}} \\\\[/tex]

r = L

[tex]\omega = \sqrt{\frac{g}{L} }[/tex]

When the mass of the ball = 65 g and length of the string = 2m, the angular velocity is calculated as follows;

[tex]\omega = \sqrt{\frac{g}{L} } \\\\\omega= \sqrt{\frac{9.8}{2} } \\\\\omega = 2.214 \ rad/s[/tex]

[tex]\omega = \frac{2.214 \ rad}{s} \times \frac{1 \ rev}{2\pi \ rad} \times \frac{60 \ s}{1\min} = 21.14 \ rpm[/tex]

Thus, the angular velocity of the ball at the given mass and length of the string is 21.14 rpm.

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