Question

can someone help me with questions 3 and 4 please !!

3. A machine part has the shape of a solid uniform sphere of mass 205 g and diameter 4.50 cm. It is spinning about a frictionless axle through its center, but at one point on its equator it is scraping against metal, resulting in a friction force of 0.0200 N at that point. Find its angular acceleration. Let the direction the sphere is spinning be the positive sense of rotation. How long will it take to decrease its rotational speed by 18.0 rad/s? Ans: -10.8 rad/s, 1.66 S 4. A 367-N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 25.0 rad/s. The radius of the wheel is 0.600 m, and its moment of inertia about its rotation axis is 0.800MRP. Friction does work on the wheel as it rolls up the hill to a stop, a height h above the bottom of the hill; this work has absolute value 2500 J. Calculate h. Ans: 13.9 m

Answer 1

3) Radius of spherical part R = 4.50/2= 2.25 cm = 0.0225m

Mass of sphere M = 205 grs= 0.205 kg

Moment of inertia I = (2/5) MR2 = (2/5) *0.205*(0.02252) = 4.15*10-5 kgm2.

Retarding Force F = 0.0200N

torque = - FR = -0. 0200*0.0225 = - 0.00045 Nm

There fore angular acceleration = torque / moment of inertia = -0. 00045/4.15*10-5 = 10.84 rad/sec2

Angular acceleration = change in angular velocity/time

Time = change in angular velocity / angular acceleration= 18/10.84= 1.66 seconds.

4)

ve et / Potential energy & workclone due to friction - Change in kinetic energy - gh & 2000 - 2 t / mul - 3674 +2500 = ² (0.8oom) (25) Given mg=367. & 2 mass m= 367 = 37.45 kg Speed - Ru= 0.600x25=ismis Radios R=0.600 m . in translational kinetic energy 2 m – 2 x 3745 (1) 4213.25 Roratiomed kinetic energy 1 x 0.800m² x po. Soox 37-45760.600.jpfer) = 33.70.55 u. 367h + 2 00= 4213. 12+ 3370-5 - 7583.62 367 h = 7583.63. e height ha 7583 62 - 20 367