#630 The 4.00-kg block in the figure isattached

The 4.00-kg block in the figure is attached - Physics

ChemistryExplain daily providing Q&A content “#630 The4.00-kg block in the figure is attached" in Physics, Ap physics 1 practice test, Best colleges for physics, Best physics books, Best physics schools

Get the Free Online Chemistry Q&A Questions And Answers with explain. To crack any examinations and Interview tests these Chemistry Questions And Answers are very useful. Here we have uploaded the Free Online Chemistry Questions. Here we are also given the all chemistry topic.

ChemistryExplain team has covered all Topics related to inorganic, organic, physical chemistry, and others So, Prepare these Chemistry Questions and Answers with Explanation Pdf.

For More Chegg Questions

• Inorganic Chemistry
• Organic Chemistry
• Physical Chemistry

Join Our Telegram Channel for Covers All Update by ChemistryExplain:- Click Now

Free Chegg Question

The 4.00-kg block in the figure is attached to a vertical rod by means of two strings. When the system rotates about the axis of the rod, the strings are extended as shown in the diagram, and the tension in the upper string is80.0 N.
A)What is the tension in the lower cord?

B)How many revolutions per minute does the system make?

C)Find the number of revolutions per minute at which the lower cord just goes slack.

Free Chegg AnswerFor More Chemistry Notes and Helpful Content Subscribe Our YouTube Chanel - Chemistry Explain  

Best Free Chegg Answer

 

We have the following forces acting on the weight

    F1    = upper string VECTOR (up and to the left)
    F2    = lower string VECTOR (down and to the left)
    W    = weight VECTOR (straight down)

The net force is the addition of these forces

    F(net) = F1 + F2 + W

By observation, the weight moves in a horizontal plane, therefore the net force is purely horizontal.

    F(net) = ( F3, 0) N

where F3 is unknown at the moment, but we expect it to be -ve, because the motion is centripetal (& therefore to the left at the instant of time of the drawing)

We need to get some results for F1, F2 and W.
Now W is the easiest,

    W = (0, -40) N   (vector is straightdown with magnitude 4N*10m/s^2 = 40N)

F1 & F2 will lie along the direction of the strings (upper and lower), and will therefore make similar triangles with that formed by the pole and the strings.

Hence from diagram we have for F1,

    x : y : r = 0.75m : 1m : 1.25m =   48N: 64N : 80N

from similar triangles OR

       F1 = (-48, 64) N   [-ve sign because it points left and up from the weight ]

& for F2 we have,

    x : y : r = 0.75m : 1m : 1.25m =  0.6*T :0.8*T : T

i.e.
       F2 = (-0.6T , -0.8T ) [-ve signs point down and to the left from the weight ]

Remembering we have,

    F(net) = F1 + F2 + W

or after substituting,

    (F3, 0) = (-48, 64) + (-0.6T , -0.8T ) + (0,-40)

A)

This gives two equations, firstly the y-component,

    0 = 64 - 0.8T + -40    => T =30N

B)

and then the x-component

    F3 = -48 -0.6T                     
    => F3 = -48 -18 = -66 N

Now F3 is a centripetal force, so

    F3 = mw^2R
    => w = sqrt(66/(4*0.75)) = 4.7 rad/s

which still needs to be converted to RPM.

C)

The tensions in the strings are going to change, but the strings will remain at the same angles. The strings are of equal length and hence cant change their positions
when under tension.

i.e. if T1 is the tension in the upper string, and T2 is thetension in the lower string we have,

       F1 = (-0.6T1 , 0.8T1 )
       F2 = (-0.6T2 , -0.8T2 )

    from similar triangles as above.

    we still have

        F(net) = F1 + F2 +W

    which gives,

    (F3, 0) = (-0.6T1, 0.8T1) + (-0.6T2 , -0.8T2 ) +(0, -40)

    when the cord goes slack (T2 =0) so,

    y-component,

        0 = 0.8T1 - 40, => T1 =50N

    x-component,

    F3 = -0.6*50N = -30N

    F3 = mw^2R
    => w = sqrt(30/(4*0.75)) = 3.2 rad/s

we'd expect that the angular speed should decrease in order for the bottom string to go slack.

Hope that helps