A uniform cylinder of mass M lies on a fixed plane inclined at an angle θ with horizontal. A light string is tied to the cylinder at the rightmost point and a mass m hangs from the string as shown. Friction between cylinder and plane is sufficiently large to prevent slipping. If the cylinder is in equilibrium for θ = 30°, the value of M = nm. The value of n is ...
Physics · Mixed Chapter
JEE Mains

Question

A uniform cylinder of mass M lies on a fixed plane inclined at an angle θ with horizontal. A light string is tied to the cylinder at the rightmost point and a mass m hangs from the string as shown. Friction between cylinder and plane is sufficiently large to prevent slipping. If the cylinder is in equilibrium for θ = 30°, the value of M = nm. The value of n is ...

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: The hanging mass m is in equilibrium, so the rope tension is T=mg. Step 2: For the cylinder, torque balance about its centre requires the contact friction magnitude f=T=mg (string acts at the tangency point with lever arm R, and friction at the point of contact has the opposite moment). Step 3: Along the incline, both weights M and m have downslope components. Force equilibrium therefore requires f=(M+m)g sinθ. Step 4: Combine mg=(M+m)g sin30°=(M+m)g/2 to obtain 2m=M+m ⇒M=m. Therefore n…