Question

A thick walled cylinder has a light string
wrapped around its outer radius and rotates
about a horizontal axis. The string then goes
vertically straight up and over a massive
pulley that also rotates about a horizontal
axis, and finally connects to a mass m =
0.900 kg on a rough incline (μk = 0.200) that
is angled at 25.0° to the horizontal.
When the system is released from rest the mass slides down the ramp a distance of 1.80 m.
a. The cylinder has an inner radius R1 = 9.00 cm and outer radius R2 = 18.00 cm and
mass MC = 0.480 kg. Calculate its moment of inertia.
b. The pulley is constructed from a hoop and 2 metal bars to form a ⨂ shape (the bars
have a length equal to the diameter of the hoop). The hoop has radius R = 12.0 cm
and mass 0.300 kg and the bars are each 0.150 kg. Find the total moment of inertia.
c. Draw a free-body diagram for all three objects.
d. What is the frictional force magnitude on the mass as it slides down the ramp?
e. Use Newton’s Laws to find the acceleration of the mass.
f. What are the angular accelerations of both the pulley and cylinder?
g. How long does it take the mass to slide down 1.80 m?
h. How many revolutions do the pulley and cylinder complete in that time?
i. Find the final speed of the mass on the incline using kinematics.
j. What is the final momentum magnitude of the mass?
k. What is the final angular momentum of both the pulley and cylinder?
l. Why is the total momentum not constant?
m. Find the final speed of the mass using energy methods and confirm your answer in
part (i).

т а

Answer 1

Soluton m 0.900 kg, AK= 0.200 Mars of ayhinder Me O480kg RI 9.00 Cm, R,- 18.00 cm Mas per anea Mc =D T(RR,) Moment ot inta of ayhinder (nik hoe) MotL of ayner (witout hale) Mo ohindr (emoves) 2 2 (TRR TR 2. MC 4 (R-R) O-L80 2 kg m 97210

Jadius of hoep R =f2.0 cm 0.300 kg 0.150 kg MASS of eaik bar Mp = MOmant of nertia of Roop+ baus Mb MR22MR 2 hb2-0)104) 0-300+ ( TO ч ка м 46.8 xro c. Fneebody diagnams FNMgcose Mgase = ma mgsMe Mg sine- F-

FF = (0.200) (0.9o0)(9-81) (a 25.0°) 60 N Ma Ff T e mg sin RA 2A-T,R 6 M(RR() 2 Since fe Atrin is not akyping, a RA = RC wb a T-T 6 (AC RR a MssMe-Ake) Mb

(0.900)(9.81) Sin 25°- 0-200 s259 9-002+ 18-02) 0-150 0 480 0.900 0300 + 397 m/s f. Pngular acc eleraton of hoop a .G44 nadls angular acc eleration ot ylnder 7.763 nad/s2 SVitat t-7 0+ 605 S a RA 15 nad= 2.387 nev R 12 J-80 ДОL - 10 xad = }.59 2 nev. 0 18

V 2as 02aL 2aL 2. 243 M/s j Fnl wementum magtde Pf= MV =2.0 186 kg mls 0 0875 kg ms I Wfh hb R. 012 kg.ws SiMe net enternal fonce Or Cystem i nt naf Conont Constant fretJ ementun i Nnk dme y gaAvilation al fone Mgisine wrnk dore by Picon Wf = -MkmgL ase WnetWf mgSine-As) 11 hange n kinetic energy

mgu (sio-Ak r0) = Vr2 Lh b 2 R 2 sin-k) 2 mgL 2 R2 2メ0.900×9.81×/-80 ( Sin5°-0.、200 Co) 25°) C46-8メro4 12-02o 97.2メ10-4 0-900 t 3.0210 Vf =2.243 ms