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Question

4)) 16% D Mon 5:10 PM a Chrome File Edit View History Bookmarks People Window Help A Home-Myusu fi Dwww.webassign.net/web/Student/Assignment-Responses/submit2dep=1 1594592 :. Apps Bookmarks A usu canvas chem sapling e chegg B Bodybuilding.com a S Austin x Ellucian Degree Works - Us x Homework 8 x Chegg Study | Guided Sol xSean White Stats, News, V x A preReq med school S slcc login USU canvas ongkick- Concert slcc canvas Other Bookmarks » ew 12.-12 points CJ8 9.FC.012 My Notes Two blocks are placed at the ends of a horizontal massless board, as in the drawing. The board is kept from rotating and rests on a support that serves as an axis of rotation. The block on the right has a mass of 3.0 kg. Determine the magnitude of the angular acceleration when the system is allowed to rotate 0.60 m 1.4 m Axis 12 kg Go Support Magnitude of the angular acceleration = | Select Read the eBook Section 9.4 Newton's Second Law for Rotational Motion About a Fixed Axis 13.-1 points CJ8 9.FC.013. My Notes The same force F is applied to the edge of two hoops. The hoops have the same mass, although the radius of the larger hoop is twice that of the smaller one. The entire mass of each hoop is concentrated at its rim, so the moment of inertia is I - Mr', where M is the mass and r is the radius. Which hoop has the greater angular acceleration, and how many times as great is it compared to that of the other hoop? 2R 26 J.

Explanation / Answer

When the board is allowed to rotate, the torques acting upwards are given by:

Sum of clockwise torque = sum of anti clock wise torque

so here each torque = F xr   

mlg*rl - mrg*rr = T

12*9.8*0.6 - 3.0*9.8*1.4 =

= 29.4 Nm

But torque is also = T = IA

where A is Alpha is the angular acceleration

I is the moment of inertia of the system

I = mlrl2 + mrrr2

I = 12*0.6*0.6 + 3.0*1.4*1.4

I = 10.2 kg m2

The angular acceleration of the system is A = T/I
  
A= 29.4/10.2

A = 2.88 rad/s2---------<<<<<<Answer

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