t Verizon 7:02 PM 1 56%- Angreen b000 with mass \"\\+2.7 kg and radius ?..0.15mh

Question

t Verizon 7:02 PM 1 56%- Angreen b000 with mass "+2.7 kg and radius ?..0.15mhangs rrom astmy that goes over bler solddel paey with mass m 2.1 kg and adis R-009.The other end of the sering is attached to a massless anel through the cencer of an orange sphere on a lat horizontal surface that rolls without slipping and hani mass m 3.6kg and radus 0.19 m. The system is releaned from rest. What is magnitude of the linear acceleration of the hoop! 2) what is magnitude of the linear acceleration of the sphere 3) What is the magnitude of the angular acceleration of the disk pulley 4 What is the magnitude of the angular acceleration of the sphere 1 What is the tension in the string between the sphere and disk pulley 61 What is the tension in the string between the hoop and disk pulley 7) The green hoop failsa distance d-1.52 m. (After being released from rest How much time does the hoop take to fall 1.52 m 8) What is the magnitude of the velocity of the green hoop after it has dropped 1.52 m 9) What is the magnitude of the final angular speed of the orange sphere lafter the green hoop han fallen the 1.52m

Explanation / Answer

Given,

mH = 2.7 kg ; rH = 0.15 m ; mD = 2.1 kg ; rD = 0.09 m ; mS = 3.6 kg ; rS = 0.19 m

1)The net force that is acting is the weight of the hoop, so

Fnet = 2.7 x 9.81 = 26.487 N

a = Fnet/M

a = Fnet/(mH + 1/2mD + 7/5mS)

a = 26.487/(2.7 + 0.5 x 2.1 + 1.4 x 3.6) = 3.01 m/s^2

Hence, a = 3.01 m/s^2

2)a = 3.01 m/s^2

3)alpha = a/R = 3.01/0.09 = 33.44 rad/s^2

Hence, alpha = 33.44 rad/s^2

4)alpha = a/R = 3.01/0.19 = 15.84 rad/s^2

alpha = 15.84 rad/s^2

5)T = Ma

T = 7/5 x 3.6 x 3.01 = 15.17 N

Hence, T = 15.17 N

6)T' = mH(g - a)

T' = 2.7 (9.81 - 3.01) = 18.36 N

Hence, T' = 18.36 N

7)t = sqrt (2d/a)

t = sqrt (2 x 1.52/3.01) = 1.01 s

Hence, t = 1.01 s

8)v = sqrt (2 a d)

v = sqrt (2 x 3.01 x 1.52) = 3.02 m/s

Hence, v = 3.02 m/s

9)w = v/R = 3.02/0.19 = 15.89 rad/s

Hence, w = 15.89 rad/s

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