A uniform solid cylinder of mass m - 2.0 kg and radius r =30 cm rotates about a

Question

A uniform solid cylinder of mass m - 2.0 kg and radius r =30 cm rotates about a fixed horizontal axis. A massless rope is wrapped around the cylinder and the end of the rope is attached to a box of mass M = 4.0 leg. If the rope does not slip on the cylinder as the mass falls, what is the magnitude of the angular acceleration of the cylinder, in rad/s^2? 14 18 22 26 30 A 500 kg mass is to be hung vertically from a 1.5-m long steel cable. The cable has a circular cross section. If we require that the cable stretch no more than 0.10 mm, what is the minimum radius of the cable, in mm? (The Young's modulus of steel is 20 times 10^10 N/m^2. Ignore the mass of the cable.] The comet Kobayashi has a perihelion distance to the Sun of 3.08 times 10^11 m and aphelion distance to the Sun of 2.27 times 10^12 m. What is its period as it revolves around the Sun, in years? [The mass of the Sun is 1.99 times 10^30 kg and one year is 3.16 times 10^7 s.]

Explanation / Answer

38) C) 22

given m = 2 kg

M = 4 kg

r = 30 cm = 0.3 m

let a is the acceleration of the block.

let T is the tension in the string.

net force on hanging block,

Fnet = M*g - T

M*a = M*g - T

T = M*g - M*a

net torque acting on the cyllinder = T*r

I*alfa = (M*g - M*a)*r

m*r^2*alfa = M*g*r - M*a*r

m*r^2*alfa = M*g*r - M*r*alfa*r

m*r^2*alfa = M*g*r - M*r^2*alfa

alfa*(m*r^2 + M*r^2) = M*g*r

alfa = M*g*r/(m*r^2 + M*r^2)

= 4*9.8*0.3/(2*0.3^2 + 4*0.3^2)

= 22 rad/s^2

39) D) 11

Apply, Y = (F/A)/(delta_L/L)

Y*(delta_L/L) = m*g/(pi*r^2)

r^2 = m*g*L/(pi*Y*delta_L)

r = sqrt(m*g*L/(pi*Y*delta_L) )

= sqrt(500*9.8*1.5/(pi*20*10^10*0.1*10^-3))

= 0.011 m

= 11 mm

40) C) 25.3 years

radius of orbit, r = (2.27*10^12 + 3.08*10^11)/2

= 1.289*10^12 m

we know, T = 2*pi*r^1.5/sqrt(G*M_sun)

= 2*pi*(1.289*10^12)^1.5/sqrt(6.67*10^-11*1.99*10^30)

= 7.98*10^8 s

= 7.98*10^8/(3.16*10^7) year

= 25.3 years

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