Question

A potter's wheel is rotating around a vertical axis through its center at a frequency of 1.7 rev/s. The wheel can be considered a uniform disk of mass 5.8 kg and diameter 0.40 m. The potter then throws a 3.1 kg chunk of clay that is shaped like a flat disk of radius 8.0 cm, onto the center of the rotating wheel. What is the frequency of the wheel after the clay sticks to it?

Answer 1

Consider the given data, Frequency of rotation, f -1.6 rev/s Mass of the disk, M -5.8 kg Diameter, D -0.4 m Radius, R 0.2m Mass of chunk, m-3.1kg Radius of flat disk, r8 cm0.08 m Solution Consider the given equation of moment of inertia of disk Substitute the value of M (mass of disk) and R (radius of disk) in equation (1) 1-(58 kg)(02m) 1-0.116 kgm2 Hence, the moment of inertia of uniform disk is [0.116kgm Here there is no extemal torque is acting , hence applying the conservation of angular momentum Consider the equation of initial angular moment L-I(2nf) Substitute the value of I and f in equation (3) l+ (0.116 kgm2)2n(1.6 rev/s) L = 1.24 kgm 2 /s Hence, the initial angular moment is 1.24 kgm/s 124 kgn Consider the equation of moment of inertia of clay Substitute the value of m (mass of chunk) and r (radius of flat disk) in equation (4)

(3.1kg) (0.08 m) 1-9.92x10- kgm Hence, the moment of inertia of the clay is 9.92x10-3 kgm2 Consider the given equation of total angular momentum Substitute the value of I and I, in equation (5) 0.116 kgm)+(992x10-* kgm') t0.126 kgm Hence the clay is chunked moment of inertia is given by 0.126 kgm Consider the equation conservation of angular momentum ω_1total tota Substitute the value of initial angular momentum and I2 in equation (6) 1.24 kgm2/:s 0.126 kgm a1 -9.85 rad's Hence the value final angular speed is 9.85 rad's Consider the equation of final frequency fi 2π Substitute the value of a, and in equation (7) (9.85 rad/s) 1.57 Hz Hence the frequency of wheel after clay stick on it is 1.57 Hz

Answer 2

Frequency of rotation,f = 1-7 rev/s Mass of the disk, m = 5.8 kg Diameter, D 0.4 m Mass of the chunk, m'-3.1 kg Radius of flat disk, r-0.08 m Here no external torque is acting, hence applying the conservation of angular momentum Initial angular momentum-final angular momentum initial angular momentum, L 10 x5.8x0.22x2x3.14x1.7 -1.24kg m1/s Moment of inertia of the clay, I'-_m'r =-×3.1× 0.082 = 9.92 x 10-3 kg m2 When the clay is chunked moment of inertia is given by, 0.116+9.92x103 -0.126 kg m From the conservation of angular momentum,

Ιω Final angular frequency, "- 1.24 0.126 -9.85rad/s o" Final frequency.f 2π 9.85 2x3.14 -1.57Hz

Answer 3

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