Question

Learning Goal:

To apply the equations of motion to a system that involves rotation about a fixed axis and to use this information to determine key characteristics.

The slender rod ABAB shown has a mass of m=53.0m=53.0 kgkg and is being supported by a rope and pulley system stationed at CC. Starting from rest (in the position shown), the rope and pulley system tug on the rod causing it to rotate about AA. The torque applied to the pulley is T=2.55T=2.55 kN⋅mkN⋅m and has an effective moment arm of r=0.110r=0.110 mm. The dimensions shown in the figure are l=2.30l=2.30 mm and h=1.30h=1.30 mm. Assume the pulley is frictionless and massless
<EC10 Equations of Motion: Rotation about a Fixed Axis 10 of 15 In Review Determine the angular acceleration of the rod the instant the rope and pulley system have pulled the rod through an angle of 8= 4.10°. Express your answer to three significant figures and include the appropriate units. Learning Goal: To apply the equations of motion to a system that involves rotation about a fixed axis and to use this information to determine key characteristics. The slender rod AB shown has a mass of m = 53.0 kg and is being supported by a rope and pulley system stationed at C. Starting from rest (in the position shown), the rope and pulley system tug on the rod causing it to rotate about A. The torque applied to the pulley is T = 2.55 kN.m and has an effective moment arm of r = 0.110 m. The dimensions shown in the figure are l = 2.30 m and h = 1.30 m. Assume the View Available Hint(s) μΑ = Value Units Figure < 1 of 1 Submit Part B - Normal Component of the Reaction at A Determine the normal component of the reaction the rod exerts on the pin at A. Use the coordinate system set up by the free-body diagram below. h B BO

Answer 1

solution: - ainen as! - Man 2.30 m = 1.30 m the rod AB (m) = 53.0kg. Torque applied to the pulley (T) = 2.55 KN.m The value of effective moment aum (1) = 0.110m The length of the dod AB(l) The vertical distance between CB ch) The angle of inclination of and (0) = 4.100 Determination the angular acceleration of the the instand the core and pulley system have pulled the dad through an angle af Q = yo leo The angular acceleration of the lod can (2) he determined as a to (m) eod - To I we know that the expression for the maus moment of inertia es as follows: I a meha - The expresion for the torque when the rod is at o = 4.100 To = fx hx Cuso 3

2 where the expunion for the dod T for face applied on F = 4 from eq. Fs 2.55 x lut 2510 Ollo 0.110 = 23 181.81 18 f 23181:82 N put 4. B into eq. ☺ To = 23101.82 x 1.30 x cus 4.100 h To 30059.2 405 N.m a A 30 a put er. ① and ę FOD q. into e. ☺ = 30059.2405 134 (2.30)2 3 30059.2 Yus 93.45666667 d = 321163.8 269/ 2 322 lad/s² T d = 322 ead/s2 Ans. PARTCB) Determinations along the normal Component of the reaction the and exents on the rin at A.

Now, the weight of the aud is mg 8 AB w h ß B AL W 53 x 9.81 = 519.93N А. Am w=519.93N) 69 Taking moment about point is [fig LG) AND Smozo (Wxf) (Anxl) =0 put all values in er. O le calculate the numal reaction (An) at pt-A- 519.93 x 2.30 .(Anx 2.30) 519-97 x 2.30 20 2 2.%0 An 2 An = 259.965 x 260 N 260 N Ans. An PART CC). Determination of the tangential component of the reaction the red enents on the pin at A

The expresion for the tangent component reaction at point A- At two calculated as - Now, tenian in the core can be t sino An= t An sino 13 put all values in ea. m t 260 3636.493618 N sin 4.10° put into es ② 3636.493610 x ces 4.200 = 3627.187041 3630N 3130 N. Ang. At = ZA At x