The 26 - in. drum rotates about a horizontal axis with a constant angular veloci
ID: 1816897 • Letter: T
Question
The 26 - in. drum rotates about a horizontal axis with a constant angular velocity Ohm = 7.5 rad/sec. The small block A has no motion relative to the drum surface as it passes the bottom position theta = 0. Determine the coefficient mus of static friction between the block and drum if the block is observed to slip as it reaches (a) theta = 50 degree and (b) theta = 100 degree. Check in the latter case to see that contact is maintained until theta = 100degree. Ans. (a) mus = 0.302, (b) mus = 0.573 Ohm = 7.5 rad/secExplanation / Answer
Friction holds the block to the drum. The component of the weight into the drum and the centripetal force together equal the normal force. The component of the weight tangent to the drum is what makes the block slip. When it just slips, the two are equal. Centripetal acceleration is always the same (as long as block isn't slipping). Component of the weight pushing in is Agcos(theta) For angles less than 90, this component will add to the centripetal force. Centripetal force is A (v^2) /r angle cos sin 50 0.64278761 0.766044443 Convert 7.5 rad/s with 13" radius to centripetal acceleration inch conversion m 13 2.54 0.3302 rad/s pi rev/s v^2/r 7.5 3.141592654 1.193662073 18.57375 Normal force = 9.8m/s^2(cos50)A + Av^2/r = N = A(6.299+18.57) Weight component parallel to drum = A9.8m/s^2(sin50)=A(7.5072) Frictional Force = uN coefficient of friction = A(7.50720)/A(6.299+18.57) = 0.3018 = 0.302 Weight// Weight perp centrip mu 7.507235543 6.299318575 18.57375 0.301821849 When the angle is 90, there is no component of the weight adding to the centripetal force, and the component parallel to the drum is the entire weight. In this case the coefficient of static friction must be 0.53 When the angle is 100, there is a component of the weight working against the centripetal force. This component will reduce the normal force, so the coefficient of friction must be larger for the block to just start to slip at 100 deg All the formulas are the same angle cos sin 100 -0.173648178 0.984807753 Weight// weight perp centrip mu 9.65111598 -1.701752141 18.57375 0.572019749 But now the perpenticular component of the weight is negative. coefficient of friction = parallel component of the weight divided by the sum of the centripetal force and the perpendicular(radial) component of the weight (which is now negative - since the denominator is smaller, the fraction is larger) mu = 0.572 (book answer has round off error) Since coefficient of friction required at 90 deg is less than what we have at 100 deg, contact will be maintained up to 100 deg with coefficient of static friction of 0.57
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