M, a solid cylinder (M 1.71 kg, R=0.129 m) pivots on a thin, fixed, frictionless

Question

M, a solid cylinder (M 1.71 kg, R=0.129 m) pivots on a thin, fixed, frictionless bearing. A string rapped around the cylinder pulls downward with a force F which equals the weight of a 0.690 kg mass i.e. F. erup theafeiade the lin guian warelwith on or the wyhindernduials the whk mass, wrapped r cet which equals the friction esa be690gmass pulls downward with a 6.769 N. Calculate the angular acceleration of the cylinder 6.14×101 rad/s^2 You are correct. Your receipt no. is 158-9184 Previous Tries If instead of the force F an actual mass m 0.690 kg is hung from the string, find the angular acceleration of the cylinder.

Explanation / Answer

Part (b)
Consider the system .

F(total)=W(weigh tof the mass) - F(due to cylinder's inertia)
or
m1a=m1g - IA/R
A=a/R (tangential acceleration of th ecylinder is the same as the mass of the weight)
m1a=m1g - .5 m2 R^2 (a/R)/R

Now we have (after cleaning up the mess);
a=(m1 g )/ (m1 + .5 m2 ) this is acceleration of the system
a=(0.690 x 9.81)/(0.690 + 0.5 x 1.71 )
a = 4.381 m/sec^2

Angular acceleration of the cylinder A
A=a/R=(m1 g )/ [R(m1 + .5 m2 ) ]
A= 4.381/0.129 = 33.96 rad/sec^2

Part (C)
h=.5at^2 (Initial velocity =0)
h2-h1=.5a(t2^2 - t1^2)=
h2-h1=0.5 x 4.381( (0.930 )^2 - (0.730 )^2)
h2-h1=0.727 m

Related Questions

Navigate

• Browse (All)
• Browse M
• Subjects

---

---

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