A block of mass m B = 140kg sitting on a frictionless ramp, where =35 degrees, a

Question

A block of mass mB = 140kg sitting on a frictionless ramp, where =35 degrees, and is attached to an idealized pulley of mass mp = 60kg and radius rp=0.15m by a weighless cord.    The pulley is spinning freely (there are no other external forces acting on it).

a. What is the angular acceleration of the pulley?

b. What is the tension in the cord?

c. If the block initially starts sliding at a height of 4.5m and an initial angular velocity of 1rad/s, how much time does it take for the block to reach the bottom of the ramp, and what's it's final angular velocity when it does?

d. How many revolutions did it go through?

Explanation / Answer

Given Data Mass of the block, mB= 140 kg Mass of the pulley, mp = 60 kg Radius of the pulley, rp= 0.15 m Angle of inclination, = 350 Solution: a) Net torque acting on the system is,                       = I                   rpT = ( mprp2 / 2 )                        T = mprp / 2                                  ...... (1) Net force acting on the block is,                   mB a = mB g sin 35 - T            mB (rp ) = mB g sin 35 - (mprp / 2)                        = 140 kg (9.8 m/s2 ) sin35 / (140 kg) (0.15 m) + [(60 kg) (0.15 m)/2]                            = 30.86 rad/s2 -------------------------------------------------------------------------------------------------------- b) From equation (1),                        T = mprp / 2                           = (60 kg) (0.15 m) (30.86 rad/s2) / 2                           =  138.87 N ------------------------------------------------------------------------------------------------------- (c) Initially, the block is at a height of h = 4.5 m,      Initial angular velocity is, i = 1 rad/s      Initial linear velocity is, vi = ri                                                = (0.15 m) (1 rad/s)                                                = 0.15 m/s      The distance travelled by the block is,                                     s = 4.5 m / sin 35                                        = 7.85 m ---------------------------------------------------------------------------------------------------------- Time taken for the block to reach the bottom is,                               s = vi t + 1/2 a t2                                        = vi t + 1/2 ( rp) t2
                        7.85 m = (0.15 m/s) t + 1/2 (30.86 rad/s) (0.15 m) t2
                       2.31 t2  + 0.15 t - 7.85 = 0                                      t = 1.81 s ---------------------------------------------------------------------------------------------------------- Final angular velocity is calculated as,                             = i + t                                = 1 rad/s + (30.86 rad/s2  )(1.81 s)                                 = 56.85 rad/s --------------------------------------------------------------------------------------------------------- d) Number of revolutions,                               = i t    + 1/2 t2                                 = (1 rad/s) (1.81 s) + 1/2 (30.86 rad/s2  )(1.81 s)2        
                                        = 52.36 rad Since , 1 rev = 2 rad             1 rad = 1 /2 rev for   52.36 rad , the number of revolutions is, = 8.3 rev          
         
                                               

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.