Chapter 9 A1 kg mass m is attached to a light cord that is wrapped around a pull

Question

Chapter 9 A1 kg mass m is attached to a light cord that is wrapped around a pulley of radius r -5 cm, which turns with negligible friction. The mass falls at a constant acce a 3 l be to estimate the moment of inertia of the pulley. Coordinate system: +2 is right; +y is down, clockwise is + for torques. 42. What force is puling the mass downward? 47. What force do the mass and the palley have in common? A) Torque, ? )Tension, T eight, mg D) Moment of inertia, I D) T 43. What force is pulling the mass upward? s put your answers for 44. and 46. using knowledge of 47, and the A) Weight, mg B) Torque, T fact that a ra. When you solve for the moment of inertia I, what "sanity check" Moment of inertia, I ension, T do you get? A) mra B) Ia C) ma 44. Apply Newton's second law EF, to the mass: A)T-m(a-g) T- mig + a) T-mig-a) 49.1 0,00326 00568 0.01068 D) 0.00174 45, What torque is causing the pulley to rotate? 50. What are the units of r? D) rmg A)J B) kg C)N 46. Apply Newton's second law for rotation 2r to the pulley

Explanation / Answer

given

m = 1 kg

r = 5 cm = 0.05 m

42) C) weight, mg

43) D) Tension, T

44) D) T = m*(g -a)

Fnety = m*g - T

m*a = m*g - T

T = m*g - m*a

= m*(g - a)

45) B) rT
Torque = r cross F

= r*F*sin(90)

= r*T

46) B) T = I*alfa/r

Torque = I*alfa

r*T = I*alfa

T = I*alfa/r

47) B) T

tension pulls the pulley downward and pulle the mass upward.

48) D) rF

49) B) 0.00568 kg.m^2

T = m*(g - a)

= 1*(9.8 - 3)

= 6.8 N

Torque = I*alfa

r*T = I*a/r

==> I = T*r^2/a

= 6.8*0.05^2/3

= 0.00567 kg.m^2

50) D) kg.m^2

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