Angreen boo with mass mh . 2.4 kg and radius Rh . 0.14 m hangs from a string tha

Question

Angreen boo with mass mh . 2.4 kg and radius Rh . 0.14 m hangs from a string that goes over a blue solid disk pulley with mass md 2.2 kg and radius Rd 0.08 m. The other end of the string is attached to a massless axel through the center of an orange sphere on a flat horizontal surface that rolls without slipping and has mass m 3.6 kg and radius R 0.22 m. The system is released from rest. What is magnitude of the linear acceleration of the hoop hWhat is magnitude of the linear acceleration of the sphere? What is the magnitude of the angular acceleration of the disk pulley What is the magnitude of the angular acceleration of the sphere? What is the tension in the string between the sphere and disk pulley "What is the tension in the string between the hoop and disk pulley h The green hoop falls a distance d-1.48 m. (After being released from rest.) How much time does the hoop take to fall 1.48 m What is the magnitude of the velocity of the green hoop after it has dropped 1.48 m? m/s What is the magnitude of the final angular speed of the orange sphere (after the green hoop has fallen the 1.48 m)? rad/s

Explanation / Answer

F net = m*g = 2.4 x 9.81 = 23.54 N

I = 7/5*mR^2

M = 7/5*m

a = F/m = 23.54/(2.4 + 1/2 x 2.2 + 7/5 x 3.6) = 2.756 m/s^2

1) linear acceleration of hoop= 2.756 m/s^2

2) linear acceleration of sphere = 2.756 m/s^2

3) w = a/R = 2.756/0.08 = 34.46 rad/s^2

4) angular acceleration sphere = a/R = 2.756/0.22 = 12.53 rad/s^2

5) tension between pulley and sphere = M*a = 7/5*3.6*2.756 = 13.89 N

6) tension between hoop and pulley = m(hoop) (g - a) = 2.4 (9.81 - 2.756) = 16.93 N

7) t = sqrt(2s/a) = sqrt(2*1.48/2.756)

t = 1.074 seconds

8) v = sqrt(2as) = sqrt(2 x 2.756 x 1.48) = 2.856 m/s

9) omega = v/R = 2.856/0.22 = 12.98 rad/s

Please rate my answer, good luck...

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