M, a solid cylinder (M=2.23 kg, R=0.127 m) pivots on a thin, fixed, frictionless

Question

M, a solid cylinder (M=2.23 kg, R=0.127 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls downward with a force F which equals the weight of a 0.870 kg mass, i.e., F = 8.535 N.

a)Calculate the angular acceleration of the cylinder.

b)If instead of the force F an actual mass m = 0.870 kg is hung from the string, find the angular acceleration of the cylinder.

c)How far does m travel downward between 0.730 s and 0.930 s after the motion begins?

d)The cylinder is changed to one with the same mass and radius, but a different moment of inertia. Starting from rest, the mass now moves a distance 0.450 m in a time of 0.470 s. Find Icm of the new cylinder. Tries 0/20 Post Discussion

M, a solid cylinder (M=2.23 kg, R=0.127 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls downward with a force F which equals the weight of a 0.870 kg mass, i.e., F = 8.535 N. a)Calculate the angular acceleration of the cylinder. b)If instead of the force F an actual mass m = 0.870 kg is hung from the string, find the angular acceleration of the cylinder. c)How far does m travel downward between 0.730 s and 0.930 s after the motion begins? d)The cylinder is changed to one with the same mass and radius, but a different moment of inertia. Starting from rest, the mass now moves a distance 0.450 m in a time of 0.470 s. Find Icm of the new cylinder. Tries 0/20 Post Discussion

Explanation / Answer

A) Tension T = force pulling the string down Torque t = T*R = I*a T = I*a/R I = 1/2 mR^2 = 1/2*2,27*0,127^2 = 0,0183064 kgm^2 T = 0,0183064*a/0,127 a = 42,8735 rad/s^2 --------------- B) And for Jim: The pulling force is still T the tension, but T = mg - ma, so mg is not equal to the tension! T = mg-ma t = TR Ia = TR I*(a/R) = TR T = I*a/R^2 mg-ma = Ia/R^2 a = mg/(I/R^2 + m) a = 6,18/(0,0183064/0,127^2 + 2,27) a = 1,81498 m/s^2 (= tangential acceleration, or acceleration of mass m) a = a/R = 1,8149/0,127 a = 14,29116 rad/s^2 = angular acceleration by mass 0,63 kg --------------- C) a = 1,8149 m/s^2 s = 1/2 at^2 s1 = 1/2*1,8149*0,63^2 s1 = 0,360167 m s2 = 1/2*1,8149*0,83^2 s2 = 0,62514 m difference d = 0,264975 m = distance moved between 0,63 and 0,83 seconds ----------- D) is this meant that the mass moves 0,2558 m from rest during the first 0,45 seconds? And what do you mean by: find Icm? s = 1/2 at^2 0,2558 = 1/2 a*0,45^2 a = 2,52 m/s a = mg/(I/R^2 + m) 2,52 = 6,18/(I/0,127^2 + 2,27) I = 0,00284142 kgm^2

Related Questions

Navigate

• Browse (All)
• Browse M
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.