A uniform disk of mass 0.75 kg and radius 5.2 cm is mounted in the center of a 1
ID: 2013193 • Letter: A
Question
A uniform disk of mass 0.75 kg and radius 5.2 cm is mounted in the center of a 10 cm axle and spun at 600 rev/min. The axle is then placed in a horizontal position with one end resting on a pivot. The other end is given an initial horizontal velocity such that the precession is smooth with no nutation.(a) What is the magnitude of the angular velocity of precession?
? rad/s
(b) What is the speed of the center of mass during the precession?
? m/s
(c) What are the magnitude and direction of the acceleration of the center of mass?
? m/s2
(d) What is the magnitude of the vertical component of the force exerted by the pivot?
? N
What is the horizontal component?
? N
Explanation / Answer
Given that mass of the dsk is m = 0.75 kg radius r = 5.2 cm disk is mounted at thecenter of a axle is D = 10cm /2 = 0.05 m a ) Magnitude of the angular velocity of precession _p = m g D / I ..........(1) moment of inertia is I = 1/2 m r^2 ........(2) substituting 2 in 1 _p = 2 g D / r^2 = 2 *9.8 m/s^2 * 0.05m / (0.052m)^2 * 600*2rad / 60s = 5.77rad /s b ) speed at the center of the mass during precession v_cm = _p * D = 5.77 rad/ s * 0.05 m = 15.4 m/s c ) Acceleration of center of mass is a_cm = D _p^2 = 0.05 m * ( 5.77rad/s)^2 = 1.66 m/s^2 d ) magnitude of the vertical component of the force exerted by the pivot is F_v = m g = 0.75 kg *9.8 m/s^2 = 7.35 N horizontal componnt is F_h = m a_cm = 0.75 kg * 1.66m/s^2 = 1.245 NRelated Questions
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