<\/a><\/a><\/p>\n

Student<\/strong>
\n3. A grinding stone in the shape of a solid disk with diameter .52 m and a mass of 50 kg is rotating at 850 rev\/min. You press an ax against the rim with a normal force of 160 N, And the grinding stone comes to rest in 7.5 seconds. Find the coefficient of friction. 4. A 15 kg bucket of water is suspended by a rope wrapped around a windlass, that is solid cylinder .3m in diameter with mass 12kg. The cylinder pivots on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls 10m to the water. You can ignore the weight of the rope. A. What is the tension in the rope while the bucket is falling? B.With what speed does the bucket strike the water? C. What is the time of fall? D. While the bucket is falling, what is the force exerted on the cylinder by the axle?<\/p>\n

Tutor<\/strong><\/p>\n

Hi! Welcome!!<\/p>\n

Student<\/strong>
\nhey! its nice to meet you. I just have a couple of problem i need help on. thanks for your help!<\/p>\n

Tutor<\/strong>
\nNice to meet you too!!<\/p>\n

ok
\nlet me see problem 3 please
\none sec please<\/p>\n

Student<\/strong>
\nok<\/p>\n

Tutor<\/strong>
\nok
\nlet me draw a diagram please<\/p>\n

Student<\/strong>
\nok!<\/p>\n

Tutor<\/strong>
\nok
\nCan you see the diagram?<\/p>\n

Student<\/strong>
\nyup!<\/p>\n

Tutor<\/strong>
\nthe ax exerts a normal force of Fn = 160 N<\/p>\n

Student<\/strong>
\nyup<\/p>\n

Tutor<\/strong>
\nok good
\nCan you remember the equation for friction force??<\/p>\n

Student<\/strong>
\nright?<\/p>\n

Tutor<\/strong>
\ncorrect!!<\/p>\n

Student<\/strong>
\nFk=Fnx(Mue sub k)<\/p>\n

Tutor<\/strong>
\nwe need to find mu sub k
\nright?<\/p>\n

Student<\/strong>
\nyup!
\nbut we don\u2019t know Fk so, we need to find that right?<\/p>\n

Tutor<\/strong>
\nexactly<\/p>\n

Student<\/strong>
\nso can i make a force diagram?<\/p>\n

Tutor<\/strong>
\nsure<\/p>\n

Student<\/strong>
\nright?<\/p>\n

Tutor<\/strong>
\nwe can assume the disk is horizontal
\nwe don’t need to consider the weight Fg
\nok?<\/p>\n

Student<\/strong>
\nright. but i just put that to balance it out.<\/p>\n

Tutor<\/strong>
\nok<\/p>\n

Student<\/strong>
\nwould Fa= (Omega) r?<\/p>\n

Tutor<\/strong>
\nwhat is Fa ?<\/p>\n

Student<\/strong>
\nFa= force applied<\/p>\n

Tutor<\/strong>
\nthe only force applied is Fn<\/p>\n

Student<\/strong>
\ni dont know that force, but there must be a counter acting force, to make Fk, and also that means there must be a constant velocity or accleration<\/p>\n

Tutor<\/strong>
\nand it makes the disk stop
\nWe can see that the only force applied is Fn
\nok?<\/p>\n

Student<\/strong>
\nok<\/p>\n

Tutor<\/strong>
\ndue to this force, a friction force Fk = u Fn appears
\nand this friction force Fk makes the disk stop
\nright?<\/p>\n

Student<\/strong>
\nright<\/p>\n

Tutor<\/strong>
\nok
\nnow we can find the angular acceleration of the disk<\/p>\n

Student<\/strong>
\nalpha= (omega) r right?<\/p>\n

Tutor<\/strong>
\ncorrect!!
\nsorry
\nlet me change something<\/p>\n

Student<\/strong>
\nso i get 23.143?<\/p>\n

Tutor<\/strong>
\nV = w r
\nand
\na = alpha r<\/p>\n

Student<\/strong>
\noh.<\/p>\n

Tutor<\/strong>
\nCan you see the eqs?
\nV = w r
\na = alpha r<\/p>\n

Student<\/strong>
\nyup!<\/p>\n

Tutor<\/strong>
\nok
\nto find alpha we need to remember a kinematics equation
\nwe know the initial angular speed
\nthe time t
\nand the final speed
\ncorrect!!
\ncorrect!!
\nwf = alpha t + wo
\nso what is alpha?<\/p>\n

Student<\/strong>
\nso Alpha=11.87
\n–
\n-11.87 rads\/sec^2<\/p>\n

Tutor<\/strong>
\nwhat is the unit?
\nit is better to work with rad\/s
\nfor angular speed
\nok?<\/p>\n

Student<\/strong>
\nok
\noh my bad, yeah thats what i got for Wo<\/p>\n

Tutor<\/strong>
\nperfect!!<\/p>\n

Student<\/strong>
\nthen i divided it by time
\nto get my alpha<\/p>\n

Tutor<\/strong>
\nalpha = -11.87 rad\/s^2<\/p>\n

Student<\/strong>
\nyes, that is what i got!<\/p>\n

Tutor<\/strong>
\ngood
\nnow
\nwe need to remember the Newton’s second law for rotations
\nfor linear motion we know Fnet = ma
\nright?<\/p>\n

Student<\/strong>
\nyup!<\/p>\n

Tutor<\/strong>
\nand for rotations we have a very similar expression
\nTorque(net) = I alpha
\nlet me write it on the board please
\nCan you see the eq on the board?<\/p>\n

Student<\/strong>
\nyes<\/p>\n

Tutor<\/strong>
\nis this expression familiar to you?
\nTnet = net torque
\nI = moment of inertia
\nalpha = angular acceleration<\/p>\n

Student<\/strong>
\nright<\/p>\n

Tutor<\/strong>
\nok good
\nwhat is I for a solid disk?<\/p>\n

Student<\/strong>
\num. 1\/2MR^2?<\/p>\n

Tutor<\/strong>
\nperfect!!!!
\nso let’s find I for the grinding stone
\nI = ?<\/p>\n

Student<\/strong>
\n1.69?<\/p>\n

Tutor<\/strong>
\nyes!!
\nI = kg m^2<\/p>\n

Student<\/strong>
\nthen i get a torque of -20.1?<\/p>\n

Tutor<\/strong>
\nyes!!
\nTnet = 20.06 N m
\nCan you remember the definition for torque?
\nthe torque applied on the disk is produced by the friction force
\nright?<\/p>\n

Student<\/strong>
\nuh yeah it makes sense, since the torque goes against omega<\/p>\n

Tutor<\/strong>
\nyes!!
\nand torque = F d
\nright?
\nlet me write it on the board
\nplease<\/p>\n

Student<\/strong>
\noh right, yeah that makes sense<\/p>\n

Tutor<\/strong>
\nok good
\nso we have the eq in magnitude<\/p>\n

Student<\/strong>
\nso Mu sub k =.483?<\/p>\n

Tutor<\/strong>
\n20.06 Nm = u Fn (0.26)
\nyes!!
\n0.48
\nGreat job on that one! Nicely done.
\nDo you have any questions about what we just did?<\/p>\n

Student<\/strong>
\nno, i think you explained it well enough, i believe i understand the problems a lot better, thank you!<\/p>\n

Tutor<\/strong>
\nyou are welcome
\nnow problem 4
\nCan you give me a moment while I take a look at problem 4??<\/p>\n

Student<\/strong>
\nok<\/p>\n

Tutor<\/strong>
\nok
\nlet me draw a diagram please
\non a new board<\/p>\n

Student<\/strong>
\nk<\/p>\n

Tutor<\/strong>
\nok can you see the diagram?<\/p>\n

Student<\/strong>
\nyup!<\/p>\n

Tutor<\/strong>
\nok good
\nwhat forces are acting on the bucket?<\/p>\n

Student<\/strong>
\ngravity, and the rope?<\/p>\n

Tutor<\/strong>
\ngood
\ngravity and tension
\nok
\nfor part A we have to find the tension
\nright?<\/p>\n

Student<\/strong>
\nyup
\nso Ft=Mbg?
\nso Ft=147N?<\/p>\n

Tutor<\/strong>
\nFg = mb g
\nFg = 15 kg x 9.8 m\/s^2
\nFg = 147 N
\nright?<\/p>\n

Student<\/strong>
\noh right<\/p>\n

Tutor<\/strong>
\nbut it is not T
\nthe bucket is falling
\nwith a certain acceleration a
\nright?<\/p>\n

Student<\/strong>
\nyes<\/p>\n

Tutor<\/strong>
\nand for the bucket we have Fnet = m a
\nor
\nFg – T = mb a
\nright?<\/p>\n

Student<\/strong>
\nyes
\nthat makes sense<\/p>\n

Tutor<\/strong>
\nok
\na is the acceleration of the bucket
\nand it is also the tangential acceleration of teh cylinder
\nright?
\ndo you agree??<\/p>\n

Student<\/strong>
\nyes, that would make sense<\/p>\n

Tutor<\/strong>
\nok
\nlet me write one more eq please
\non the white board<\/p>\n

Student<\/strong>
\nk<\/p>\n

Tutor<\/strong>
\nwhat is the moment of inertia of a solid cylinder?<\/p>\n

Student<\/strong>
\ni believe the same as the disk problem<\/p>\n

Tutor<\/strong>
\ngood
\n1\/2 m r^2<\/p>\n

Student<\/strong>
\nso .135 Alpha= Tr<\/p>\n

Tutor<\/strong>
\nok good
\nand alpha = a\/r
\nright?<\/p>\n

Student<\/strong>
\nright<\/p>\n

Tutor<\/strong>
\nok good
\nI labeled equations 1 and 2
\ncan you see it??<\/p>\n

Student<\/strong>
\nyup<\/p>\n

Tutor<\/strong>
\nwe can take these two eqs
\nand form a system of equations
\nright?<\/p>\n

Student<\/strong>
\nyup<\/p>\n

Tutor<\/strong>
\nwe have two eqs
\nand two unknown
\nvariables
\na and T<\/p>\n

Student<\/strong>
\nso should we plug in for T or a?<\/p>\n

Tutor<\/strong>
\nwe need both<\/p>\n

Student<\/strong>
\ni think a right?<\/p>\n

Tutor<\/strong>
\nyes
\nany is ok<\/p>\n

Student<\/strong>
\nso i simplified it. and got a=T\/6
\nis that right?><\/p>\n

Tutor<\/strong>
\nperfect!!
\nnow let’s replace a in eq 1<\/p>\n

Student<\/strong>
\nso i get T=42?<\/p>\n

Tutor<\/strong>
\nperefct!!
\nT = 42 N
\ngood
\nis it clear so far?<\/p>\n

Student<\/strong>
\nyeah, things are clearing up, thanks a bunch!<\/p>\n

Tutor<\/strong>
\nnote that we are using the same equations we used in the previous problem
\nTnet = I alpha
\nT = F d
\nand
\na = alpha r
\nright?<\/p>\n

Student<\/strong>
\nyeah!<\/p>\n

Tutor<\/strong>
\nand the Newton\u2019s second law
\nFnet = ma
\nok good
\nlet’s see part B
\nB.\u00a0 With what speed does the bucket strike the water?
\nwe can find a now
\nright?
\nwhat is a?<\/p>\n

Student<\/strong>
\n7m\/s^2<\/p>\n

Tutor<\/strong>
\nyes a= 7 m\/s^2
\nnow we have a kinematics problem
\nthe bucket starts from rest
\nand falls 10 m
\nwhat is the final velocity?
\nlet me see<\/p>\n

perfect!!
\nVf = 11.83 m\/s<\/p>\n

Student<\/strong>
\nand that means the time to hit the water is 1.69?<\/p>\n

Tutor<\/strong>
\nyes!!
\nt = 1.69 s<\/p>\n

Student<\/strong>
\nok part D. this one is more complicated
\nwould it be 147N?<\/p>\n

Tutor<\/strong>
\nit would be the same magnitude as tension
\nright?
\nthe cylinder is rotating
\nbut it is in translational equilibrium<\/p>\n

Student<\/strong>
\nso it would not be 42N?<\/p>\n

Tutor<\/strong>
\nT = 42 N<\/p>\n

Student<\/strong>
\nright. but that wouldnt be the force on the cylinder?<\/p>\n

Tutor<\/strong>
\nyes
\nit would be the force on teh cylinder applied by the axle<\/p>\n

 <\/p>\n

Student<\/strong>
\nok so how do we find this force?<\/strong><\/p>\n

Tutor<\/strong>
\nlet me draw a diagram please
\nCan you see the diagram?
\nthe force F is exerted by teh axle
\non the cylinder
\nright?<\/p>\n

 <\/p>\n

Student<\/strong>
\nyeah that makes sense<\/p>\n

Tutor<\/strong>
\nthe cylinder is in translational equilibrium
\nso the equilibrium conditions must be verified
\nSum Fx = 0
\nSum Fy =0
\nSum Fy = F – T = 0
\nright?<\/p>\n

Student<\/strong>
\nright that makes sense
\nand there is no force in the X direction right?<\/p>\n

Tutor<\/strong>
\nexactly
\nso F = T<\/p>\n

Student<\/strong>
\nso F=42?<\/p>\n

Tutor<\/strong>
\ngood<\/p>\n

Student<\/strong>
\nso it is 42?<\/p>\n

Tutor<\/strong>
\nexactly
\nYou got it!<\/p>\n

Student<\/strong>
\nhaha. thanks.<\/p>\n

Tutor<\/strong>
\nDo you have any questions about what we just did?<\/p>\n

Student<\/strong>
\nno i think i got the jist of it. thank you so much!<\/p>\n

Tutor<\/strong>
\nyou are welcome
\nThanks.<\/p>\n

Student<\/strong>
\nhave a good day!
\nColaborador: Luis Rodr\u00edguez, Tutor Boliviano de F\u00edsica<\/span><\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"

Student 3. A grinding stone in the shape of a solid disk with diameter .52 m and a mass of 50 kg is rotating at 850 rev\/min. You press an ax against the rim with a normal force of 160 N, And the grinding stone comes to rest in 7.5 seconds. Find the coefficient of…<\/p>\n","protected":false},"author":9,"featured_media":3616,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"gallery","meta":[],"categories":[89,87],"tags":[],"post_series":[],"_links":{"self":[{"href":"https:\/\/dev.latinhire.com\/es\/wp-json\/wp\/v2\/posts\/3590"}],"collection":[{"href":"https:\/\/dev.latinhire.com\/es\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/dev.latinhire.com\/es\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/dev.latinhire.com\/es\/wp-json\/wp\/v2\/users\/9"}],"replies":[{"embeddable":true,"href":"https:\/\/dev.latinhire.com\/es\/wp-json\/wp\/v2\/comments?post=3590"}],"version-history":[{"count":1,"href":"https:\/\/dev.latinhire.com\/es\/wp-json\/wp\/v2\/posts\/3590\/revisions"}],"predecessor-version":[{"id":3617,"href":"https:\/\/dev.latinhire.com\/es\/wp-json\/wp\/v2\/posts\/3590\/revisions\/3617"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/dev.latinhire.com\/es\/wp-json\/wp\/v2\/media\/3616"}],"wp:attachment":[{"href":"https:\/\/dev.latinhire.com\/es\/wp-json\/wp\/v2\/media?parent=3590"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/dev.latinhire.com\/es\/wp-json\/wp\/v2\/categories?post=3590"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/dev.latinhire.com\/es\/wp-json\/wp\/v2\/tags?post=3590"},{"taxonomy":"post_series","embeddable":true,"href":"https:\/\/dev.latinhire.com\/es\/wp-json\/wp\/v2\/post_series?post=3590"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
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