13. (20 points) A hovering bug of mass m= 0.07 kg drops from above and lands at the center of a turntable of mass M = 0.
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13. (20 points) A hovering bug of mass m= 0.07 kg drops from above and lands at the center of a turntable of mass M = 0.
Now suppose the bug crawls toward the rim of the turntable as shown in refer Figure 10(C), (e) What is the moment of inertia of the system when the bug reaches the rim of the turntable? (f) What is the angular speed (₂) of the system when the bug reaches the rim of the turntable?
Relevant equations & conversions . Symbols have their usual meanings 1 revolution - 2 radians 1 minute 60 seconds; 1 hour 60 minutes 1 kg-1000 g: 1 mi-1609 m; 1 m-3.28 ft; 1 m - 100 cm Magnitude of the acceleration due to gravity: g = 9.8 m/s². Relevant equations Roots of the quadratic equation Ax² + Bx+C =0x • Area of a triangle of base band height : Abi • Area of a rectangle of length I and width w: A=Iw. . . Circumference of a circle of radius r. C 2mtr. •Area of a circular disk of radius r: A r². Trigonometric functions Newton's second law of motion Η ΣΕ mid OR E (2Fs = mas -B+√B²-4AC 24 F. Bairs MUCA N Figure 1 ΣΕ, « May Weight of an object of mass m: to mg (magnitude). Hooke's law for the restoring force exerted by a spring) (Fp)-kAx, where k is the spring constant and Aris the displacement from the equilibrium position. Frictional forces - →Magnitude of the maximum force of static friction: fimas, where n is the magnitude of the normal force acting on the body and p. is the coefficient of static friction. Magnitude of the force of kinetic friction: fun, where is the coefficient of kinetic friction. Uniform circular motion: The acceleration of an object moving in a circle of radius r at a constant speed pointing toward the center of the circle) Impulse and momentum Linear momentum of an object of mass m and velocity : M Impulse of a force = area under the Force-lime curve Fag At Impulse-momentum principle for Newton's second law): Ap Tor A Law of conservation of momentum: The total momentum of a system an isolated system is a constant. Le su EF At. to P1+ P2 + of
Work and Energy Work done by a constant external force F exerted on a system while that system undergoes a displacement d W=Fd cos 6 where F is the magnitude of the force, d is the magnitude of the displacement, and 6 is the angle between the vector F and the vector d. Kinetic energy: K= m². • Gravitational potential energy: Ug=mgy, Elastic potential energy: U₁=kx², where x is the displacement of the spring from its equilibrium (or unstretched) position (x = 0). • Work-Energy equation states that K+U+U+AEth = K₁+U₂ + Uu+W where W is the work done by all the external forces acting on the system and AE, is the change in thermal energy of the system. • The increase in thermal energy of the system due to friction is AE. Wl, where W/ is the absolute value of the work done by the force of kinetic friction.. Rotational dynamics →The torque exerted by a force F: Tr F, where r, is the moment arm (or lever arm). →For an object in equilibrium: Er=0 and 2F-0 # →The moment of inertia of a system of particles: I-Emr mr + mar} + mar] + .. →The moment of inertia of various objects: →For a single particle of mass m in circular motion of radius r: 1m² →For a solid disk of mass M and radius R rotating about an axis passing through its center: MR² →For a hoop (or ring) of mass M and radius R rotating about an axis passing through its center: 1 = MR² →Newton's second law for rotational motion: Erla, where I is the moment of inertia of the rotating object. →Angular momentum or rotational momentum: L = la If the net torque acting on a rotating rigid body is zero, i.e., ET=0, the the angular momentum of the rigid body does not change, i.e., L-la-constant or huyu.