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# sol10 - HW Solutions 10 8.01 MIT Prof Kowalski Angular...

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HW Solutions # 10 - 8.01 MIT - Prof. Kowalski Angular Energy and Angular Momentum . 1 ) 10.70 Please refer to figure 10.54 p.399. The general strategy to solve this problem is to figure out the torque around a point and the direction of the torque will tell you which way it rotates. Choose the z direction to be perpendicular to the plane shown in the figure and pointing up . If the direction of the torque is up it means that it rotates counterclockwise or it moves to the left . If the direction of the torque is down it means that it rotates clockwise or it moves to the right . The smartest choice of the point you want to write the torque about is the contact point with the ground because the torque due to friction and normal force is Zero around this point since they are acting at this point. Weight vector is also passing through this point so it doesn’t generate any torque around it. In the following calculation regard the contact point as origin: −→ r contact point = 0 a ) F to right; bottom of axle (# 1 ): −→ τ c = −→ r F c × −→ F ˆ z (Where c denotes the contact point; −→ τ c : torque around the contact point c ; −→ r F c : −→ r to force from contact point) So it rotates to the right. Now if you write the torque around center of mass the only force that has torque around it is −→ f . Consider z component of −→ τ cm about center of mass. To roll to right, torque must be in ˆ z , since F gives a +ˆ z torque, f (friction) must give ˆ z torque hence points to left . 1

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F to right; top of axle (# 2 ): −→ τ c = −→ r F c × −→ F ˆ z So it rotates to the right. With the same argument friction can point to the left(if axle radius is small) or to the right (if axle radius is large).
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