Question

In the figure below, the hanging object has a mass of m, = 0.480 kg; the sliding block has a mass of m, = 0.825 kg; and the pulley is a hollow cylinder with a mass of M = 0.350 kg, an inner radius of R4 = 0.020 0 m, and an outer radius of R, = 0.030 0 m. Assume the mass of the spokes is negligible. The coefficient of kinetic friction between the block and the horizontal surface is My = 0.250. The pulley turns without friction on its axle. The light cord does not stretch and does not slip on the pulley. The block has a velocity of V, = 0.820 m/s toward the pulley when it passes a reference point on the table. R R mg (a) Use energy methods to predict its speed after it has moved to a second point, 0.700 m away. m/s (b) Find the angular speed of the pulley at the same moment. rad/s

Answer 1

Aş both the blocks are whnected by the same cood, so they will have same velocity 'V. angellar velocity 'w' of the pulley is given by W = v R2 also moment of inertia q pulley is given by I M CR?+R}) = 0.350%10-02 2460-03%gma? = 2.275x10 + Kg -m? From the work energy theorem, APEM, = AKEM + ARE ma tko + Ake Pulley OPEM, = mg(n-o) = ngre = 0.480x 9.8x0.75 = 3.29 J OkEm = 1 m, 1² m, v? AkEma = 1 m2 v2 - 1 mzu? = 'AKE, take me = h (m, tm, ) v2 - 1 (o tome)? = (0.6525 v² - 0.4387) J

и : Кто, хи - Да, и - (0-25 xo 825x38x 0 10) т 2 1.4IS T ARE Pulleya II (wh- w, ?) - 3 : у AKGalley = 275615 9 : 0-1264 (v2 - to 82 % ) : 0- 1244 v2 - Dogs

=) 3.29 - (0.6525v2-0.4387) 41.415 + (0.12% (AV2 -0.085) =) 3.29 20:7789 v² + 0.8913

- 0.7789 v2 = 3.29-0.8913 22.40 v2 = 2.40 -3.08 07789 Vz53.08 mida 1.755 mis ta y 2 1755 radis = 58.5 m/s b2

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