I need help with this. Consider the familiar arrangement of masses shown below.
ID: 1450846 • Letter: I
Question
I need help with this.
Consider the familiar arrangement of masses shown below. This lime the pulley has nonzero mass W_g and radius r. Use the following values in this problem: M = 2.94 kg. m. = 1.2 kg. r = 0.2 m, and the moment of inertia of the pulley is Squareroot = M_pr^2/2. There is no friction at the axle of the pulley about which it routes, hut friction keeps the cord in tight contact with the outer edge of the pulley (no slipping between the cord and the pulley). If the hanging mass falls from rest through a distance of 0.25 m, find the angular velocity of the pulley at that instant. rad/s (plusminus 0.01 rad/s) Now we will find the acceleration of the hanging mass. If you do this with Newton's second law you will need 3 equations: F = Ma for the upper mass, F = ma for the hanging mass, and r = la for the pulley, plus the corrections between angular motion and linear motion. This is possible, but it's completed. Here's an easier way to do it. (You can read about this in Section 9.6 of the Chapter Summaries in the subsection about using energy to find the acceleration.) In part (a) you. no doubt, wrote down an energy conservation equation involving mass speed v and falling distance y (or something like it). Change your formula for the total energy E of the system so that each kinetic energy term has v in it (not omega). If you were careful with directions (positive y pointing down is a good choice) it will be true that v = dy/dt for each of the moving masses. After making sure that this is true, differentiate your energy formula with respect to time and set it to zero (energy is a constant). Use dy/dt = nu and dy/dt(y^2) = 2nu(dy/dt) =2nua, where a is the acceleration of the having mass. You should find that v cancels out of the equation dE/dt = 0 so that you have a single equation that you can use to solve for the acceleration a. The acceleration that you find this way is: (plusminus0.01) Now that you have the acceleration you can use F = ma on the hanging mass to find the value of tic tension T_1: ~N (plusminus 0.01 ~N) Now use Newton's second law again to find T_2 ~N (plusminus 0.01 ~N)Explanation / Answer
a)
h = distance fallen = 0.25 m
Ip = moment of inertia of pulley = mp r2/2 = (0.5) (0.2)2/2 = 0.01 kgm2
using conservation of energy
PE of hanging mass "m" = Kinetic energy of mass "m" + KE of mass "M" + rotational KE of pulley mp
mgh = (0.5) m V2 + (0.5) M V2 + (0.5) Ip w2
(1.2) (9.8) (0.25) = (0.5) (2.94) V2 + (0.5) (1.2) V2 + (0.5) (0.01) (V/0.2)2
V = 1.16 m/s
w = V/r = 1.16 / 0.2 = 5.8 rad/s
b)
for the hanging block , force equation is given as
mg - T1 = ma
1.2 x 9.8 - T1 = 1.2 a
T1 = 11.76 - 1.2 a eq-1
for the other block , force equation is given as
T2 = Ma
T2 = 2.94 a eq-2
for the pulley , torque equation is given as
T1 - T2 = I(a/r2)
using eq-1 and eq-2
(11.76 - 1.2 a ) - 2.94 a = 0.01 a /(0.2)2
a = 2.7 m/s2
c)
using eq-1
T1 = 11.76 - 1.2 a = 11.76 - 1.2 x 2.7 = 8.52 N
d)
using eq-2
T2 = 2.94 a
T2 = 2.94 x 2.7 = 7.94 N