Two equal masses m are attached by a string. One mass lies at radial distance r from the center of a horizontal turntable which rotates with constant angular velocity ω , while the second hangs from t

Question

Two equal masses m are attached by a string. One mass lies at radial distance r from the center of a horizontal turntable which rotates with constant angular velocity $ω$ , while the second hangs from the strings inside the turntable's hollow spindle (see figure). The coefficient of static friction between the turntable and the mass lying on it is $μ_{s} = 0.5$ . The maximum and minimum values $r_{max}, r_{min}$ of r such that the mass lying on the turntable does not slide. Find the ratio of $r_{max} / r_{min}$ ?

TestHub · Accepted Answer

For r min ​ case, friction will act in the outward direction and will be of maximum value. From F.B.D.: T = m g T − μ m g = m ω 2 r min ​ m g − μ m g = m ω 2 r min ​ ⇒ r min ​ = ω 2 g ( 1 − μ ) ​ For r ma x ​ case, friction will act in the inward direction and will be of maximum value. T = m g m g + μ m g = m ω 2 r ma x ​ ⇒ r ma x ​ = ω 2 g ( 1 + μ ) ​ Now: r min ​ r ma x ​ ​ = ω 2 g ( 1 + μ ) ​ × g ( 1 − μ ) ω 2 ​ = 0.5 1.5 ​ = 3 r min ​ r ma x ​ ​ = 3

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