A uniform solid sphere with mass M and radius 2R rests on a horizontal tabletop.

Question

A uniform solid sphere with mass M and radius 2R

rests on a horizontal tabletop. A string is attached by a

yoke to a frictionless axle through the centre of the

sphere so that the sphere can rotate about the axle. The

string runs over a disk-shaped pulley with mass M and

radius R that is mounted on a frictionless axle through

its centre. A block of mass M is suspended from the

free end of the string. The string does not slip over the

pulley surface, and the sphere rolls without slipping on

the tabletop.

(a) The system is released from rest. Find an expression for the speed v of the block, after it

has fallen a distance h. Your expression should involve only numbers, g, and h. Show and

explain your work.

(b) Use kinematics to find the acceleration of the block. (Check your answer: a = 10g/29.)

A uniform solid sphere with mass M and radius 2R rests on a horizontal tabletop. A string is attached by a yoke to a frictionless axle through the centre of the sphere so that the sphere can rotate about the axle. The string runs over a disk-shaped pulley with mass M and radius R that is mounted on a frictionless axle through its centre. A block of mass M is suspended from the free end of the string. The string does not slip over the pulley surface, and the sphere rolls without slipping on the tabletop. The system is released from rest. Find an expression for the speed v of the block, after it has fallen a distance h. Your expression should involve only numbers, g, and h. Show and explain your work. Use kinematics to find the acceleration of the block.

Explanation / Answer

a)

M g h = 0.5 M v^2 + 0.5 I1 w1^2 + 0.5 I2 w2^2 + 0.5 M v^2

I1 = (2/5) M (2R)^2 = 1.6 M R^2

I2 =0.5 M R^2

w1 = v/(2R)

w2 = v/R

M g h = 0.5 M v^2 + 0.5 I1 w1^2 + 0.5 I2 w2^2 + 0.5 M v^2

M g h = 0.5 M v^2 + 0.5 (1.6 M R^2) (v/2R)^2 + 0.5 (0.5 M R^2) (v/R)^2 + 0.5 M v^2

cancel M:

g h = 0.5 v^2 + 0.5 (1.6 R^2) (v/2R)^2 + 0.5 (0.5 R^2) (v/R)^2 + 0.5 v^2

==> 2 g h = v^2 + 0.4 v^2 + 0.5 v^2 + v^2

==> 2 g h = (1 + 0.4 + 0.5 + 1) v^2

==> 2 g h = (2.9) v^2

==> 20 g h = (29) v^2

==> v = sqrt((20/29) g h)

b)

vf^2 - vi^2 = 2 a h

==> v^2 - 0^2 = 2 a h

==> (20/29) g h = 2 a h

==> a = 10g/29

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