An green hoop with mass mh = 2.8 kg and radius Rh = 0.17 m hangs from a string that goes over a blue solid disk pulley with mass md = 2 kg and radius Rd = 0.09 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 ms = 4 kg and radius Rs = 0.2 m. The system is released from rest.
1)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?
5)What is the tension in the string between the sphere and disk pulley?
6)What is the tension in the string between the hoop and disk pulley?
7)The green hoop falls a distance d = 1.56 m. (After being released from rest.) How much time does the hoop take to fall 1.56 m?
8)What is the magnitude of the velocity of the green hoop after it has dropped 1.56 m?
9)What is the magnitude of the final angular speed of the orange sphere (after the green hoop has fallen the 1.56 m)?
mechanics Physics question!?
2011-10-31 2:38 pm
回答 (3)
2011-11-01 12:54 am
✔ 最佳答案
Hellothe only force on the system is the weight of the hoop
F net = 2.8kg*9.81m/s^2 = 27.468 N
The inert masses that must be accelerated are the ring, pulley, and the rolling sphere.
The mass equivalent of M the pulley is found by
torque τ = F*R = I*α = I*a/R
F = M*a = I*a/R^2 -->
M = I/R^2 = 21/2*m*R^2/R^2 = 1/2*m
The mass equivalent of the rolling sphere is found by:
the sphere rotates around the contact point with the table.
So using the theorem of parallel axes, the moment of inertia of the sphere
is I = 2/5*mR^2 for rotation about the center of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere.
I = 7/5*mR^2
M = 7/5*m
the acceleration is then
a = F/m = 27.468/(2.8 + 1/2*2 + 7/5*4) = 27.468/9.4 = 2.922 m/s^2
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1)linear acceleration of hoop= 2.922 m/s^2
2) lin. accel. of sphere = 2.922 m/s
3) angular acceleration disk pulley:
α = a/R = 2.922/0.09 = 32.46 rad/s^2
4) ang. acceleration sphere = a/R = 2.922/0.2 = 14.61 rad/s^2
5) tension between pulley and sphere = M*a = 7/5*4*2.922 = 16.3632 N
6) tension between hoop and pulley = m(hoop) (g - a)
= 2.8(9.81 - 2.922) = 19.29 N
7) s = 1/2*at^2 --> t = √(2s/a) = √(2*1,56/2.922)
--> t = 1.068 seconds
8) v = √(2as) = √(2*2.922*1.59) = 3.05 m/s
9) ω = v/R = 3.05/0.2 = 15.24 rad/s
Regards
2015-03-17 4:25 am
8) v = √(2as) = √(2*2.922*1.59) = 3.05 m/s 1.59 should be 1.56
2014-10-28 7:18 pm
Thank you so much
收錄日期: 2021-04-13 18:20:00
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