Need some help wth the formulas and steps on this one please. 1). A small partic

Question

Need some help wth the formulas and steps on this one please.

1). A small particle of mass m is pulled to the top of a frictionless half-cylinder (of radius R) by a cord that passes over the top of the cylinder as illustrated in the figure below. (a) Assuming the particle moves at a constant speed, show that F = m gcos(?). Note: If the particle moves at constant speed, the component of its acceleration tangent to the cylinder must be zero at all times. (Do this on paper. Your instructor may ask you to turn in this work.) (b) By directly integrating W =

Explanation / Answer

Since the direction of the tangent line changes, the direction of the force along the tangent line also changes, so that the integral formula for the work done by the tension in the cord must be used … W = ? (v_F ) • ( v_dL ) … where … v_F is the force vector describing the tension in the cord and v_dL is the displacement vector along the semicircle … since both ( v_F ) and ( v_dL ) are tangent to the semicircle at any time, the angle ß between them is zero ( ß = 0 ), so that … … W = ? (v_F ) • ( v_dL ) = ? F ( dL ) cos ß = ? F dL … where … dL = R d ?, the arc length of the displacement along the semicircle … therefore … … W = R ? F d ? … to know the tension F in the cord, look at all the forces acting on the particle … aside from the tension F in the cord, the other forces acting on the particle are its weight W acting vertically downward and the normal force N exerted by the cylinder surface acting upward along the radius … since the particle moves at constant speed, there is no acceleration and the vector sum of three forces acting must be zero … ( v_F ) + ( v_W ) + ( v_N ) = 0 … or … ( v_F ) = - ( v_N ) - ( v_W ) … where the force vector ( v_F ) is purely tangential while the force vector ( v_N ) is purely radial … expressing the forces in terms of the radial (denoted by the subscript 1) and tangential (denoted by the subscript 2) components … we get the following … ( v_F ) = F2 e2 … … ( v_N ) = N1 e1 … ( v_W ) = - ( mg sin ? ) e1 - ( mg cos ? ) e2 … where … … e1 = unit vector along the radial direction … … e2 = unit vector along the tangential direction … it follows that … … F2 e2 = - N1 e1 - [ - ( mg sin ? ) e1 - ( mg cos ? ) e2 ] ……….. = ( mg sin ? - N1 ) e1 + ( mg cos ? ) e2 … comparing the coefficients … … F1 = 0 since F is purely tangential … --> … mg sin ? - N1 … or … N1 = mg sin ? … F2 = mg cos ? … that is, the magnitude F of the force vector ( v_F ) is … F = F2 = mg cos ? … the work done by the force is therefore … … W = R ? F d ? = R ? (mg cos ? ) d ? = m g R ? cos ? d ? = m g R sin ? … Finally, evaluating the result of the integral from ? = 0 to ? = p / 2 , we get … … W = m g R

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